Does this summation identity hold?

Question

Suppose we have $\sum_{n=-\infty}^{-N} f(n)$. Is this sum equal to $\sum_{n=N}^{\infty} f(-n)$?

The reason why I ask this question is because I am trying to write $\sum_{- \infty}^{\infty} \frac{(-1)^n}{(2n-1)^3}$ as a summation of form $\sum_1^{\infty} $. Am I correct in my reasoning?

score 2 · Accepted Answer · answered Nov 28 '15 at 15:48

2

Yes, you have correctly rewritten that sum. Your sum can therefore be rewritten as $$ \sum_{n=-\infty}^\infty \frac{(-1)^n}{(2n-1)^3} = -1 + \sum_{n=1}^\infty \frac{(-1)^n}{(2n-1)^3} + \sum_{n=1}^\infty \frac{(-1)^{-n}}{(2(-n)-1)^3} $$

answered Nov 28 '15 at 15:48

Ben Grossmann

225,327

but isnt $(-1)^{-1} = -1 $ ?? – Nov 28 '15 at 16:02
Indeed, so you could simplify that last sum quite a bit – Ben Grossmann Nov 28 '15 at 16:03
so, can we conclude that $(-1)^{-n} = 1 ? $ – Nov 28 '15 at 18:13
No, we can't. Try a few values of $n$ to check. – Ben Grossmann Nov 28 '15 at 20:57

Does this summation identity hold?

1 Answers1