They are not only natural

Number Theory Level 3

What is sum of all the integers ??

none of these undefined 1 0

2 solutions

Vishal S
Feb 24, 2015

I think my solution is correct.If it is wrong, sorry for my mistake

$\infty$ = $\frac {1}{0}$

- $\infty$ =- $\frac {1}{0}$

Therefore sum of - $\infty$ + $\infty$ =- $\frac {1}{0}$ + $\frac {1}{0}$ = $\frac {0}{0}$ =undefined

Ok, this is just plain wrong! Arithmetic operations on undefined values is mathematically incorrect!

Prasun Biswas - 6 years, 3 months ago

Yes Prasun is right !!!!! We cannot perform ARITHMETIC operations on undefined quantities !!!!! So it all depends on how one looks at the problem from his point of view BUT technically speaking the sum equals $0$ .

Gagan Raj - 6 years, 3 months ago

Doesn't the Riemann Sum claim that sum of positive integers is -1/12?

Devin Ky - 6 years, 1 month ago

sir, if you say $\frac{0}{0}$ is undefined then $\frac{1}{0}$ is also undefined only no? can you give a proof? note: donot think in another manner that i am teasing you sir if you thoughts go like that then i feel sorry sir

sudoku subbu - 6 years, 3 months ago

Gagan Raj
Feb 23, 2015

Shouldnt it be zero technically ????? Because its not given sum of all +ve integers rather its given sum of all integers irrespective of wether they are +ve or -ve..............so its supposed to be zero right ????? Because all -ves cancel out with all +ves............so which means the answer is supposed to be zero right????

I'm confused ?????

@Calvin Lin Sir do you have any comments on this one ?????

Please Help!!!!!

what is (-infinity) + (+infinity) ?? Is it zero ??

Vighnesh Raut - 6 years, 3 months ago

Technically, the sum that you mentioned can be seen as a telescoping series and so some people say that the sum is $\boxed{0}$ . The sum can be rewritten in two ways:

$\sum_{i=-\infty}^\infty i=\sum_{i=0}^\infty \left(i+(-i)\right)=\sum_{i=0}^\infty (0)=\boxed{0}$

$\sum_{i=-\infty}^\infty i=\sum_{i=-\infty}^0 i + \sum_{i=0}^\infty i = \boxed{\textrm{ undefined }}$

The first case modifies the sum and makes it a telescoping one where all the values gets cancelled. The second case sums the positive and negative parts individually which turns out to be divergent. There's no proper answer to this problem in my opinion. Although, if one asks the limiting value of the sum, we can say that the limit is finite and equal to $\boxed{0}$ .

And if I recall correctly, infinite sums are technically sums where the upper/lower limits tends to infinity (doesn't reach infinity).

Prasun Biswas - 6 years, 3 months ago

$\sum _{ i=-\infty }^{ \infty }{ i\quad } \\ =\quad \sum _{ i=0 }^{ \infty }{ i } -\sum _{ i=0 }^{ \infty }{ i } \\ =\quad undefined\quad (as\quad we\quad could\quad not\quad subtract\quad two\quad infinities)$ ..... Same mistake was done by me in this problem.... Check my solution there , I made a mistake of subtracting two infinities ...

Vighnesh Raut - 6 years, 3 months ago

@Vighnesh Raut – Then again, one can interpret your problem as the following limiting sum:

$S=\lim_{n\to \infty}\left(\sum_{i=-n}^n i\right)\\ \implies S=\lim_{n\to \infty} \left(\frac{n(n+1)}{2}-\frac{n(n+1)}{2}\right)=\lim_{n\to \infty} 0 = \boxed{0}$

Prasun Biswas - 6 years, 3 months ago

Yes even i had the same thought in my mind. The answer is supposed to be zero but as mentioned the answer can also be equal to infinity when viewed from a different point of view. It all depends on the way of thinking of the solver. But can anyone tell me WHAT THE ACTUAL ANSWER IS ?????

Gagan Raj - 6 years, 3 months ago

but isn't that because there are different Infinities? these are exactly the same but opposite signs

Josh Kari - 6 years, 3 months ago

thats what iam asking what is -x+x ? note: x may be any value

sudoku subbu - 6 years, 3 months ago

They are not only natural

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