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What is the period of this function? And how do you come about your answer?

$$\epsilon_n(x)=\sum_{\mu=-\infty}^{+\infty}(x+\mu)^{-n}$$

I understand that $\epsilon_n(x+k)=\epsilon_n(x)$ (sort of)

Hugh Entwistle
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1 Answers1

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Hint. By the change of index $\nu=\mu\pm 1$ one gets $$ \epsilon_n(x)=\sum_{\mu=-\infty}^{+\infty}(x+\mu)^{-n}=\sum_{\nu=-\infty}^{+\infty}(x\pm1+\nu)^{-n}=\epsilon_n(x\pm 1), $$ one may prove the period is equal to $1$.

Olivier Oloa
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  • I'm a bit confused by this - when you say change the index: $\nu=\mu\pm1$

    Wouldn't this result in: $$\epsilon_n(x)=\sum_{\mu=-\infty}^{+\infty} (x + \nu \pm 1)^{-n}=\epsilon_n(x\pm1)$$

    Where does the $1$ go in your first substitution? (I am not very familiar with this topic so apologies if I am being extremely ignorant here)

     – Hugh Entwistle Aug 22 '16 at 08:34
  • @HughEntwistle Sure. This was a typo. Edited. Thanks. – Olivier Oloa Aug 22 '16 at 08:43