Quadruple Sum
S = i = 1 ∑ ∞ j = 1 ∑ ∞ k = 1 ∑ ∞ l = 1 ∑ ∞ i + j + k + l ( − 1 ) i + j + k + l = lo g ( a ) − b a a , b are co-prime positive integers. Find a + b .
NOTE : S does converge.
The answer is 5.
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3 solutions
Your first step isn't justified. n = 4 ∑ ∞ n ( − 1 ) n ( 3 n − 1 ) doesn't converge .
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True,the sum doesn't converge according to W|A,but what I am doing in the above procedure gave me the correct answer. Reason for problem: the procedure I did above is valid for ∣ x ∣ < 1 ,but I have used x = − 1 ,maybe I took limit.But,I don't know why it isn't justified.
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While doing what you have done, the order of summation got changed (the terms got rearranged). This is a conditionally convergent sum, and rearranging isn't allowed. Check this out.
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@Pratik Shastri – True,you are right,but isn't this sum symmetric.
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@Akshay Bodhare – Yeah it is symmetric. But changing the order of the terms isn't allowed. You can change the order of the summation signs if you want, but that'll be somewhat useless.
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@Pratik Shastri – But,isn't the answer correct,so what went wrong which got right again in the end?
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@Akshay Bodhare – I don't know. I guess something weird must have happened when you put x = − 1 , which isn't allowed..
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@Pratik Shastri – But,I think that the limit is justified,right?
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@Akshay Bodhare – Can you please explain first line , I am little bit confused.Thank you
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@U Z – I think you know the possible number of integer solutions using the stars and bars logic.
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@Akshay Bodhare – Yes. getting it
@Akshay Bodhare – I like a lot your avatar :)
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@Héctor Andrés Parra Vega – It's Eren Jaeger from Attack On Titan.
I had done the same too but after Pratik's clarification, I will read the suggested page in more depth.
Well this some doesn't converge.
Well, I guess that the problem is that while doing what you have done, the order of summation got changed (the terms got rearranged). This is a conditionally convergent sum, and rearranging isn't allowed.
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But why rearranging is not allowed.
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Check this out.
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@Pratik Shastri – So can you please post your solution so that I may see another method of solving this. Please, this is a nice problem.
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@Ronak Agarwal – The trick here is to write i + j + k + l ( − 1 ) i + j + k + l as − ∫ 0 1 ( − x ) i + j + k + l − 1 d x and the rest follows easily.
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@Pratik Shastri – Hi Ronak and Pratik , what is the first thing to start after our JEE syllabus to learn the basics of all the problems you do.I mean order of summation,Riemann series theorem , zeta function , residue theorem. I searched for them on wikipedia but most of things I could'nt understand,(not good mind) ( in each line there is a link) , so it's better that you and Ronak suggets me a link which contains just basics of these things.Thank you
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@U Z – Hey,if you really want to learn complex analysis so much,a friend of mine said you should start by reading walter rudin and alfohrs.
@U Z – You can check out the tons of questions on Mathematics StackExchange. It's a brilliant resource.
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@Pratik Shastri – Yes it's true , but I can't sit 24 hrs in front of computer , can I ask what was the first thing you studied about complex analysis , its sure that you startred with basics , what was that basic?
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@U Z – I haven't studied complex analysis yet, though I want to.
@U Z – Yes me too, I too have not studied complex analysis.
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@Ronak Agarwal – How do you know about zeta functions , order of summation? You too from Mathematics stack exchange? Thank you
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@U Z – Pratik might say that he doesn't know complex analysis , but he knows everything else , just name it . From Fourier Transform to Poisson's summation , Z and Laplace Transforms everything !! In fact I have learned quite a lot from him , Gamma , Beta , Alpha , Fourier series and the list goes on and on !!
@U Z – Learning about zeta function is just a matter of definition, I read the definition from wikipedia and have been using it since then, similar for gamma function and similar for beta function.
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@Ronak Agarwal – BTW the proof u post of ° Zeta funtion ; its amazing.....
@Ronak Agarwal – @Ronak Agarwal
Hi Ronak this may not be the right place to discuss it , but after reading Hriday's post , I couldn't help myself .
I recently came up with a different proof for ζ ( 2 ) = 6 π 2 .The application of that method has been used in framing my latest question , which coincidentally Pratik has solved , so I'm guessing either he solved using my method or he has an even better one !!
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@A Former Brilliant Member – Can you please name your question, please.
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@Ronak Agarwal – Geometry and Calculus go hand in hand .
@A Former Brilliant Member – Well can you please tell me what do you mean by :
"but after reading Hriday's post, I coudn't help myself"
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@Ronak Agarwal – Haha what I had actually meant was that this is supposed to be a place for discussing solutions of Pratik(the poster) ,but after seeing Hriday mentioning about your note on ζ ( 2 ) , I thought that I might as well as mention the fact that I had worked on another proof here rather than in your note .
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@A Former Brilliant Member – Then you should mention your proof as a note.
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@Ronak Agarwal – It's too long btw, I think I'll do it after Jee Mains . Also I was considering that there might be other proofs for ζ ( 2 ) here on Brilliant , so is it possible if any moderator or the staff , you know compile and post it as a single note .
Just a thought .
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@A Former Brilliant Member – Longer than mine.
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@Ronak Agarwal – Yeah , I first had to prove that sin θ = θ ⋅ r = 1 ∏ ∞ ( 1 − r 2 π 2 θ 2 ) , this was the longer part , rest is routine .
EDIT * I just noticed your other discussion, it's basically the same proof !!
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@A Former Brilliant Member – So that theorem is valid !! I was playing around with sin θ = 2 sin 2 θ cos 2 θ when I got that theorem .
Cool !!!
@A Former Brilliant Member – What other discussion. And what theorom you are saying as valid.
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@Ronak Agarwal – The discussion of Pratik's note to which you had provided the link to .
Actually the formula of expressing sin θ as a product of infinte terms , was something I came up while fiddling with the formula of sin \2 θ so I didn't know how to verify it , hence I ended up posting a question on it .
But after reading that other discussion (Link provided above) I found that it is indeed valid .
Hey !!! Where did u bring such type of QUESTIONS ; it's nice...... PLEASE TELL ME ITS SOURCE IF U want t0 tell.....
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I'd seen a similar double summation somewhere, I don't remember where though. I'll inform you when I do remember.
Don't need to complicate people, just find the number of times a term appears in the sequence, in this case, term number n appears as much times as it can be represented by ordered quadruples which sum n which entries at least 1 ((n-1)C3) then, after this, remember the identity for power series to natural logarithm; in order to obtain the three consecutive numbers that multiply the binomial you should divide in first place by x to have it levated to the (n-1), after that derive three times each side of the identity and finally multiply by -1 and divide by 6 to get the binomial, the result is our dear log(2)-(2/3) evaluating by one. (Detailed in the older post jaja, still i wanted to comment :P)
Let, i + j + k + l = n ,This happens ( 3 n − 1 ) times.
Thus,sum reduces to, ∑ n = 4 ∞ n ( − 1 ) ( n ) ( 3 n − 1 ) = ∑ n = 3 ∞ n + 1 ( − 1 ) ( n + 1 ) ( 3 n ) .
For,calculating this sum,Consider,
∑ n = 3 ∞ x n = 1 − x x 3 Differentiate thrice,then multiply by x 3 and then integrate once,in the end put x = − 1 .Divide the result by 6.
Answer is lo g 2 − 3 2