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− 1 + 2 − 3 + 4 − 5 + ⋯ = ?
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12 solutions
You are very near to answer but answer is -1/4...... You have taken 1-1+1-1..... = 0 which is not right.....As per mathametician Grandi's proof it is 1/2.....so you need to take 1/2....check my proof I have explained in detail.....
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Well the Grandi series itself doesn't converge. I posted the solution just to show how equations can be manipulate but you should not arrive at erroneous conclusions from it.
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Explanation for taking 1/2 Now, lets go further, you are absolutely right Grandi series does not converge. It accumulate either at 1 or 0 at alternate stage. So, I should take either 1 or 0. But I cannot take either 1 or 0. You can accept 1 or 0 when you are talking in terms of n(nth step), but you cannot take either 0 or 1 if you are talking in terms of \infty, because in terms of infinity nth step does not exist so you cannot take either 0 or 1. Now, if you imagine if it is not at 1 or 0 so where it is? It will be at 1/2 at any given step of nth, that is most appropriate point(1/2) if we want to correlate in terms of infinity. Please take in consideration that infinity and nth step is totally different thing. So, this is my explanation for taking 1/2 for Grandi's series. Further Explanation on current Question Hence the series given in this question is also goes to \infty, hence if I take consideration of evaluated partial sum, that is not appropriate, because evaluated partial sum is 'what is at nth step', but we cannot take nth step consideration same as explained for 0 and 1 for Grandi series above. Why I cannot take \infty or -\infty because if you are considering one of them you can not deny other one so, if you are taking +\infty you cannot deny -\infty. Here also we are ending up in a situation like either should I take 1 or 0 same as Grandi's situation. So we can not take both. Now question arises so where it is in terms of infinity, now if you could have imagined 1, 0 and 1/2. You can imagine for this also it is at -1/4, you can alway find it at -1/4.
In the second step, shouldn't it be T+1? Just curious.
If you are manipulating... (-1+2)+(-3+4)+(-5+6)...=1+1+1+1... But, strangely enough, -1+(2-3)+(4-5)+(6-7)...=-1+-1+-1+... One of the 'tricks' my analysis teachers told me is basically nondivergent sums can do that. If you can do that with a sum, show 2 different rearrangements that equal different sums, the sum itself must not converge.
By rearranging the sums in the following way it can be seen that the infinite sum is infinity.
-1 -3 -5 .... +2 +4 +6 .....
+1 +1 +1+.....
Thus the sum is infinity. There is no other answer.
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My thought exactly.
The accepted answer exceeds infinity so it is impossible.
Or we pair up adjacent numbers so we have (-1+2)+(-3+4)+(-5+6)+... which is just 1+1+1+... which is infinity.
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This is exactly what I did
But then you're ignoring odd numbered sequences, which will always be negative. Like: 1-3, 1+1-5, 1+1+1-7, 1+1+1+...-infinity, etc
If you add T to itself, getting T + T =1+-1+1+-1+.....1+-1+1+-1+1+-1.....= T
So, according to your "manipulation", 2T = T Or 1 = 1/2
There is a fallacy in the second step, you have added -1 to the first side then you can add only one -1 to the other side. But you're going on adding -1 to all the integers. This solution is flawed.
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Thanks. I have edited the answer.
Given the other options of 0 , 1 , − ∞ , and ∞ the answer is that "None of the other answers are correct."
The sequence, − 1 + 2 − 3 + 4 − 5 + . . . , does not converge. The limit of the partial sums also does not approach either ∞ or − ∞ . Consider:
| # of Terms | Unevaluated Sum | Evaluated Partial Sum |
| 1 | − 1 | − 1 |
| 2 | − 1 + 2 | + 1 |
| 3 | − 1 + 2 − 3 | − 2 |
| 4 | − 1 + 2 − 3 + 4 | + 2 |
| 5 | − 1 + 2 − 3 + 4 − 5 | − 3 |
| 6 | − 1 + 2 − 3 + 4 − 5 + 6 | + 3 |
| 7 | − 1 + 2 − 3 + 4 − 5 + 6 − 7 | − 4 |
| 8 | − 1 + 2 − 3 + 4 − 5 + 6 − 7 + 8 | + 4 |
The partial sums of this series get arbitrarily large ( → + ∞ ) and arbitrarily small ( → − ∞ ) .
I have a question though... what if we tried another approach towards solving the problem? We pair up consecutive terms such as -1 and 2, -3 and 4, -5 and 6.... and add the numbers in each pair so we get 1,1,1,1... which amounts to an infinite number of 1s which give us the sum as ∞ . Is there any fallacy in this method of solving the problem? This sort of comparison has been used by people like George Gamov who illustrated it in his novel one, two, three... infinity and he also claimed that the infinity of odd numbers is equal to the infinity of even numbers. Is there any mistake with solving the problem this way or is this also an alternate solution?Please let me know!
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This is a great question. (I don't think I or anyone else is really going to be able to do it full justice in this thread - I think it would actually make a great Brilliant wiki.)
An attempt at an answer:
When the partial sums of a series don't converge on a value, we say that the series does not converge to a sum. This is why:
Shashwat Bhardwaj and Bob Oaks are using the grouping:
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Pedro Carvalho suggests:
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Both solutions are only using the associative property of addition to group the terms differently, but because the partial sums of the series don't converge, different groupings can make the partial sums as negative or as positive as you'd like.
And you can do even stranger things if you start using the commutative property of addition to move terms around and break some terms up before regrouping. Effectively, this is what Harsh Khatri is getting at in his comment which begins with an "algebraic proof" that the series adds up to 4 1 and if you change the proof around just a bit, you can get that the series equals + 1 0 0 or any other value you'd like.
In short, the normal rules for algebra, including the properties that we depend on addition and equivalency having, can't be applied to the sums of infinite sets which don't converge.
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Aha! I see! So if a series doesn't converge, manipulating it with normal algebra isn't right because we can essentially manipulate it to get a different sum if we use a different approach right? Thanks a lot! This really clears everything up!
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@Shashwat Bhardwaj – Excellent! :) And, yes, in this case, because the sum not only doesn't converge but also fluctuates between arbitrarily high and low values, that's why you're able to use just the associative property to make it look like both − ∞ and + ∞ are plausible answers. There are also some series which don't converge but for which you cannot pull exactly this trick, only similar ones like the manipulation that Harsh Khatri uses to get 4 1 as an 'answer.'
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@Zandra Vinegar – @Zandra Vinegar , Can you check the summation method for the series of 1-2+3-4+5........and proof for that at Link for series 1-2+3-4+5..... This series is also divergence series. Fluctuate arbitrarily from high and low value. And there is no trick used for explaining that these series has a value of 1/4. And your series is same as above link just minus sign common so it will end up as -[1-2+3-4+5.......], so will not have the value of -1/4 when done using Euler & Borel/Abel summation method? Or can you claim that this all explanation of 1-2+3-4+5......is invalid in the link.
@Zandra Vinegar – Actually none of the answers are correct, and that can be explained by Riemann Rearrangement theorem
More than splendid and eye catching solution /comment /remark !
Kindly check my answer I have proved that and sidhharth also got the same answer....Though I have given complete proof, if still doubt there I can explain in detail.....
Okay, now what if you skip the -1 and pair up every consecutive term after it? 2 - 3 is -1, 4 - 5 is -1, 6 - 7 is -1, and so on, which amounts to an infinite number of -1s.
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Doesn't ignoring part of the problem negate any further explanation?
You could bit-shift the pairings and then it becomes - infinity (from the infinite number of -1 pairings), -1 (the first number which didn't get into a pairing) And + infinity (the last number which didn't get into a pairing)
The sum of which: - inf - 1 + inf = -1 altogether, assuming the infinity is the last number in the sequence, and the second to last is infinity minus one and so on and so forth,
Well now I think of it, it won't be an infinite number of pairings, it'll be (infinity/2) - 1 pairings, so it's
-1[(infinity/2) - 1] - 1 + infinity
Which if you expand becomes
- infinity/2 + 1 - 1 + infinity
Ones cancel, so answer would be infinity - infinity/2 = infinity / 2
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The concept of infinity itself is that it is unbelievably large. It means that you simply can't count it. Even if you tried counting the pairs you'd never stop, meaning that for all practical purposes it is infinity. In other words, infinity/2=infinity. Besides, this method is invalid. See Zandra Vinegar's solution-it is the right one.
This was my approach too.
Yes..I did it the same way !!
I think you have not understand the real meaning of infinity. Infinity is not a number it is not finite and not a number at all so we can divide like that . I too exactly thought like you at first.
That was exactly my approach
That mean the answer is zero or what ?
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There is no right answer, check out Zandra Vinegar's solution.
u are right..but a small mistake..there wont be infinite no. of 1 but infinite/2 because when u pair up and get a 1..u actually get the value '1' for every two no.
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The concept of infinity itself is that it is unbelievably large. It means that you simply can't count it. Even if you tried counting the pairs you'd never stop, meaning that for all practical purposes it is infinity. In other words, infinity/2=infinity
Yes, that is correct, but I think this works similarly to Grandi's series, which there is a wiki on.
It has definit answer, it is -1/4...please check my detail proof....
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I agree with Zandra Vinegar that the series does not converge, but Imran Ansari's explanation is very interesting and very insightful into a unique nature of number series. Good!
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Hi Ramanath, there are proof that divergence series can be assigned a value. I have just updated my answer and in the second section have provided the link for that article where they have proved that divergence series 1-2+3-4+5......has a value 1/4 and our question is -[1-2+3-4+5.....] which is minus sign common, so will have the value of minus sign, hence -1/4.
I have a question. if we we rewrite this as -1 -3 -5 -7 -9 ... +2+4+6+8... giving Σ1-2n +2n = 1 . where is my error? I do accept I am wrong by the way.
Hi, I have a question. If we start with -1, the answer will always be positive answer. If we start with 1-2+3-4...., the answer will always be negative answer. But is there any way we can have a 0 answer for this kind of equation (not positive or negative)? Thanks!
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This is a little tricky to answer, since what you're asking is basically, "can we make it look like the answer is 0 using reasonable-looking but fallacious reasoning?" In short, yes. Very frequently in math, if you accept a single false axiom, like "1=-1" then you can use it to prove anything you want in combination with other valid axioms. Here are two fairly convincing ways to make it seem like the sum of the series should be 0:
Let
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Therefore, since
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Alternatively, you'll probably bamboozle more people with the argument that:
2 S = − 2 + 4 − 6 + 8 − 1 0 + 1 2 . . . = ( − 2 + 4 ) + ( − 6 + 8 ) + ( − 1 0 + 1 2 ) . . . = 2 + 2 + 2 + 2 . . . and S = − 1 + 2 − 3 + 4 − 5 + 6 − 7 + 8 − . . . = ( − 1 + 2 − 3 + 4 ) + ( − 5 + 6 − 7 + 8 ) + . . . = 2 + 2 + 2 + 2
Therefore, since 2 S = S , subtract S from both sides to get S = 0 .
Take away: you can use this trick or similar ones to make the series look like it equals anything you want. That's why we say the series is non-convergent and has no numerical sum.
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How would you disprove it being equal to zero?
Infinity is infinity and dose not have a sign. the opposite of infinity is zero not minus infinity. Therefore any thing that tends to +inf. or - inf. simply tends to inf. (can provide detail if required) therefor this tends to infinity.
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That's nonsense. Positive and negative infinity are not the same. You don't get positive infinity by adding up negative numbers.
Grammatically, should be “None of the other answers is correct”.
(-1+2)+(-3+4)+(-4+5)+(-5+6)+...= 1+1+1+1+1... = inf.
THE ANSWER IS -1/4 and I can prove that.......Let me give you some history, Italian mathematician Grandi in year 1703 published and proved that -1+1-1+1-1......= 1/2 and other mathematician prove that 1+2+3......= -1/12 and these both sequences are used in many area of theories and particularly string theory. You will not accept second one just looking at there but both has a solid proof and has been accepted by mathematical society and used in several modern theories.......So now come to our equation...to prove our equation I have taken help of proof of above two theories.......PROOF------------- Let, S = -1+2-3+4-5+6........ And add same S to S but the second S shift to one step aside.......................................... S = -1 + 2 - 3 + 4 - 5............................ +S= _ -1 + 2 -3 + 4 - 5............................. 2S = -1 + 1 -1 + 1 -1+........ Now 1-1+1-1+1........ = 1/2 as per Grandi proof.... So, 2S = -1 + 1/2....... If you solve above question..... S = -1/4........
@Harsh Khatri has pointed out very important thing that Grandi's series does not Converge and that point is very TRUE and I appreciate its point. But want to put further Explanation of taking 1/2 for Grandi's series and taking -1/4 for this Question series..
Explanation for taking 1/2 Now, lets go further, you are absolutely right Grandi series does not converge. It accumulate either at 1 or 0 at alternate stage. So, I should take either 1 or 0. But I cannot take either 1 or 0. You can accept 1 or 0 when you are talking in terms of n(nth step), but you cannot take either 0 or 1 if you are talking in terms of infinity, because in terms of infinity nth step does not exist so you cannot take either 0 or 1. Now, if you imagine if it is not at 1 or 0 so where it is? It will be at 1/2 at any given step of nth, that is most appropriate point(1/2) if we want to correlate in terms of infinity. Please take in consideration that infinity and nth step is totally different thing. So, this is my explanation for taking 1/2 for Grandi's series.
Further Explanation on current Question Hence the series given in this question is also goes to infinity, hence if I take consideration of evaluated partial sum, that is not appropriate, because evaluated partial sum is 'what is at nth step', but we cannot take nth step consideration same as explained for 0 and 1 for Grandi series above. Now as @Zandra Vinegar pointed out "The partial sums of this series get arbitrarily large +Infinity and arbitrarily small -Infinity". Now let me explain why we cannot take +infinity or -infinity because if you are considering one of them you can not deny other one, So if you are taking +infinity you cannot deny infinity. Here also we are ending up in a situation like either should I take 1 or 0 same as Grandi's situation. So we can not take both. Now question arises so where it is in terms of infinity, now if you could have imagined 1, 0 and 1/2. You can imagine for this also it is at -1/4, you can alway find it at -1/4.
PROOF FOR THE DIVERGENCE SERIES VALUE
In the previous part I tries to prove using Grandi series, as several people pointed out about divergence series and it cannot be prove but in the following mentioned link they have given the summation method for the divergence series and current question can easily proved.
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In the following link of Wikipedia they have given the proof for the divergence series of 1-2+3-4+5....... where they have proved 1/4 value for this series, kindly check abel summation method and Euler and Borel proof of divergence series.
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The divergence series in this question is -1+2-3+4...... which is - [1-2+3-4+5.....] which is minus sign common hence using abel summation method and Euler & Borel method it can be proved same way.
link for proof of 1-2+3-4........
In the above mentioned article on the FIRST PAGE they have proved that DIVERGENCE series 1-2+3-4+5...... HAS A VALUE 1/4. and same way our question which is the same form of that equation with minus sign common -[-1+2-3+4-5+6.........] can also be proved that this divergence series has a value of -1/4.
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I understand your proof, but when I searched for Grandi's series,I found a Wikipedia page which claimed that though it is possible to obtain 1/2 using some proofs, the series in reality is a diverging series (Zandra Vinegar talked about this in the first answer) and therefore it doesn't have an actual sum. Here's the link... ( Grandi's Series )
Generally when we talk about infinite sums, the equals (=) sign is meant to indicate convergence to a particular value. The series is divergent, so there is no numerical answer to problem. It is however possible to assign values to divergent sums (as with Grandi's series) using alternative methods (Grandi's uses Cesaro summation), but you should understand that this is not the same as the sum being "equal" to -1/4. You should always specify the form of summation you mean when you assign value to divergent series - in any case, it is not the form of summation implied by the question.
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Hi, your point is absolutely right, Can you check the following link to wikipedia, where they have explain for the 1-2+3-4+5...... series and please check the abel summation method Euler and Borel summation method, they arrived the result at 1/4....and under the title of Stability and linearity they have proved this series is equal to 1/4 without using Grandi series and in middle paragraph they have mentioned about appropriate summation method for these kind of divergence series and doing summation of that. So, the whole article is about 1-2+3-4..... and our Question is - [ 1-2+3-4........] which is same with - sign out hence for 1-2+3-4..... = 1/4(plus) and for -1+2-3+4..... = -1/4(minus) link for proof of 1-2+3-4........
I'm not an expert on the alternate definitions of summations that are being referenced (Cesàro and Abel summations are the two at the top of the first article you link to). But here is my intuition for the bigger picture:
The most relevant experience that I have comes from game theory and from John Conway's book "Of Numbers and Games" in particular. One fascinating section of this book proves that each different way that you can setup a game called Hackenbush, corresponds with a surreal number. (The surreal numbers are the largest possible ordered field of numbers.) The amazing part is that you can use the connection between games and numbers which Conway rigorously defines, to predict the game play of large games of Hackenbush which are made from combining two or more small games. Therefore, in a very cool and useful way, the games behave like the surreal numbers.
I think all of the varieties of summation alternatives being discussed now are creating useful ways to valuate these diverging series in a similar way to how Hackenbush games can be associated with surreal numbers. In particular, it's important to note that these alternate 'summations' are not literally sums of the series, but rather, each kind of sum provides a specific technique for assigning values to many infinite, diverging series of a particular form. I think, in most cases, this is done by considerably restricting if and how terms can be reordered and regrouped before summation.
I don't have enough experience to speak on the practical motivation for these value-assigning techniques. What I do find amazing though, is that you can use these "numerical tags" to compare series that are similar enough in form to warrant the same valuation/summation technique being applied. Also, the variation between the different techniques (aka Cesàro vs. Abel etc.) is fascinating, in my opinion! :)
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@Zandra Vinegar, I am truly thankful for your honest answer and your explanation on your experience on "Of Numbers and Games".
I would like to make very humble and kind request as follow. These all summation methods are particularly and purely developed to prove the pure numeric theories and has been accepted by the mathematical societies and some has practical use. Now anybody can say which series is divergence and non-divergence and arrive at some conclusion by observing few steps of partial sum of series, in your first comment of explanation you arrived at -Infinity and +Infinity, till that step it is absolutely right, no doubt. And mathematician knows very well to that point. But these all theories and summation methods and proofs has been developed by mathematician to go beyond that and have been proved hence we can not deny that facts and it needs to be referenced when we talk on that series or about numbers. Hence you need to make reference of all that answers and methods which has been proved in your first comment of explanation along with your view and your explanation. Because if professional mathematician comes on brilliant.org, he will find it totally back-dated. If all references is given, and it has to be, then they will think that "Ok author has referred all the aspect in the area hence he/she knows what it is all about and along with this he/she is giving his/her view, then your view will be appreciated". Because everybody's view is important either it is erroneous or true.
About this question, your conclusion of -Infinity to +Infinity is very true, but several people came to the answer of -1/4, because they have referred other advanced proved theories.
Hence, since you are a author, and as you are authorized to make final conclusion, (we are just come up with our own answers here), hence, you need to refer all possible things and need to provide in your final conclusion along with your own view. Very Thanks
In a geometric series: (First term) / 1 - (common ratio)
If absolute value of the common ratio is greater or equal to one limit does not exist (MEAN GIRLS WOOO)
mate, where's the geometric series?
By the way, the ratio test you are describing works for all SEQUENCES (a series is a sum of all the terms in a sequence), except for when the ratio equals 1, when dodgy stuff happens (it could be either convergent or divergent).
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+1,-1,+1,-1...is a geometric series of ratio -1. Hence the summation formula +1/(1-(-1)) = 1/2. But I don't think it can be applied when abs(ratio) >= 1
This is the second time my answer is already filled in before I get a chance to pick an answer. I not going to play if this keeps happening.
what is MEAN GIRLS WOO?
We let S be the series such that...
S = − 1 + 2 − 3 + 4 − 5 + . . .
Notice if we add the series to itself and shift the terms as we add them we get...
S + S = − 1 + ( 2 − 1 ) + ( − 3 + 2 ) + ( 4 − 3 ) + ( − 5 + 4 ) . . .
2 S = − 1 + 1 − 1 + 1 − 1 . . .
Let T be the series such that...
T = − 1 + 1 − 1 + 1 − 1 . . .
We added T to itself as we did to S and see...
T + T = − 1 + ( 1 − 1 ) + ( − 1 + 1 ) + ( 1 − 1 ) + ( − 1 + 1 ) . . .
2 T = − 1
T = − 1 / 2
Now solving for S...
2 S = T
2 S = − 1 / 2
S = − 1 / 4
Since this is none of the answers, the answer is not here!
We know ( 1 + x ) − 2 = 1 − 2 x + 3 x 2 − 4 x 3 + . . . . . . . Put x = 1 and multiply − 1 both side. − 4 1 = − 1 + 2 − 3 + 4 . . . . . . . Hence answer is − 4 1
great solution!
I am not sure but I think that your solution is not perfectly correct because the expansion you have used is for ∣ x ∣ < 1 I am not sure whether it holds for x = 1 ...Let me know what is correct ..Please !! @Abhinav Sinha
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Why is (-1+2) + (-3+4) + (-5+6) + ... = 1 + 1 + 1 + 1 + ... = infinity not correct?
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You might pair the numbers in another way also as
− 1 + ( 2 − 3 ) + ( 4 − 5 ) + . . . . = − ∞
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@Ankit Kumar Jain – @Calvin Lin @Pi Han Goh Sir , please contribute to the discussion to make me clear as to what the actual solution to the problem is.
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@Ankit Kumar Jain – Read up Riemann Series Theorem .
What this solution writer did was to find the Abel's sum of this series, which was not explicitly stated in the question, so he is wrong.
very good and intelligent solution!
why would it become a fraction? we are dealing with whole numbers.
It can be found by considering it as the analytic continuation of the Riemann zeta function. Then you get the answer -1/4, as others here have also explained.
Here is how to do it in Mathematica: Sum[(-1)^n n^-s, {n, 1, Infinity}] /. s -> -1
None of the other answers are correct. Minus one quarter.
We can consider the sum as that of numeric suite terms with Un = (-1)^(n+1) (n+1) (∑Ui)0<=i<=n depends on wether n is even or odd
If n is odd, it exists an integer k / n=2k+1 (∑Ui)0<=i<=n = k+1
If n is even, it exists an integer p / n=2p (∑Ui)0<=i<=n = -(p+1)
So the absolute value of sum will tend to ∞. But the value will fluctuate between -∞ and +∞
A question. By finding infinite sum of an Arithmetico-Geometric Progression, taking A=0, R=-1 and D=1, I'm getting answer -1/4. People in the thread are getting diffrent answers. What is the flaw in my method?
Yes, there is a flaw. The very basic requirement for the convergence an infinite sum is ∣ r ∣ < 1 .
While here, you have substituted r = − 1 and hence the formula is not valid here.
Cant you use the alternating harmonic series? S = 1-1/2+1/3-1/4 ... = ln(2) -1/S = the series above, so the solution must be -1/ln(2)
Depends on your definition of infinity. If infinity equals all possible values in creation, the answer is infinity. If one arbitrarily assigns a sign to infinity, I suppose one could then say it's +/- infinity. But since it's entirely arbitrary to assign any specific value or range of values to infinity, +/- infinity equals infinity, as does any other infinite subdivision of infinity. Therefore, the answer is Infinity.
Siddharth you answer is right, but I can not get how you get.....I have just posted complete proof of that.......
Seems to me that the partial sum S(2n)=n and S(2n-1)=-n. So the series does not converge.
First see why infinity or negative infinity is wrong...!!!!! Since there are infinite negative odd numbers and infinite positive even numbers, the sum would be (infinity - infinity), which is indeterminate....so, we will have to find the sum... S=-1+2-3+4-5+6-7+..... =-(1+3+5+7+9+...)+(2+4+6+8+....) =-n²+2(1+2+3+4+....). [Where n= is any no.] =- n²+2(m²+m)/2. [Where m=another no.] =-n²+m²+m [Hence, none of the answers are correct]
Let the sum be T
T = − 1 + 2 − 3 + 4 − . . . .
add T and T as follows
T = − 1 + 2 − 3 + 4 . . .
T = . 0 − 1 + 2 − 3 + 4 . . .
Adding 0 and -1,-1 and +2 , +2 and -3 and so on to get
2 T = − 1 + 1 − 1 + 1 − . . . . .
As we don't know that number of one's is odd or even the series would converge to 0.5
2 T = 0 . 5
T = 0 . 2 5
adding the first pair of terms you get (-1+2)=1 adding the second pair of terms you get (-3+4)=1 and so on… you will get a and infinite sum of ones that tends to infinity
Though the sequence does not converge, we can still play with the equations.
Consider T = 1 + − 1 + 1 + − 1 …
T − 1 = − 1 + 1 + − 1 + 1 + − 1 …
T + T − 1 = 0
T = 2 1
Now, S = − 1 + 2 + − 3 + 4 …
S + T = ( − 1 + 1 ) + ( 2 + − 1 ) + ( − 3 + 1 ) + ( 4 + − 1 ) + …
⇒ S + T = 1 + − 2 + 3 + − 4 + …
⇒ S + T = − S
⇒ 2 S = − T
⇒ S = − 4 1
This is just a manipulation. In reality, the absolute value of the sum tends to ∞ while the value of the sum keeps fluctuating between ∞ and − ∞ .