What is an identity for $$\sum_{n=1}^\infty \frac{1}{k^n}\quad ?$$ I found numerous identities for $\sum_{n=1}^\infty ak^n$, all of which extremely complex, but are there any simpler identities?

  • $\begingroup$ not a good edit, who is m? $\endgroup$
    – user67133
    Apr 19 '13 at 21:46
  • $\begingroup$ @user67133 Fixed, thanks :) $\endgroup$
    – KevinOrr
    Apr 19 '13 at 21:47
  • $\begingroup$ the usage of k,n in the summation was not conventional. I modified to what I think should be the more conventional way, please review. $\endgroup$
    – jimjim
    Apr 19 '13 at 22:50
  • 1
    $\begingroup$ @Arjang: since there is an answer already using the original notation, let's leave it. $\endgroup$
    – robjohn
    Apr 19 '13 at 23:13
  • $\begingroup$ @KevinOrr : Google geometric series $\endgroup$ Apr 19 '13 at 23:18

The formula is extremely simple: when $|k|<1$, $$\sum_{n=1}^\infty ak^n=\frac{ak}{1-k}.$$ In particular, if you let $\ell=\frac{1}{k}$, then $$\sum_{n=1}^\infty\frac{1}{k^n}=\sum_{n=1}^\infty\ell^n,$$ and now apply the first identity. Note that you need $|\ell|<1$ in order for the first identity to hold, and that this is equivalent to $|k|>1$.

  • $\begingroup$ You've mixed up all the variables so that they have (1) nothing to do with the original question, (2) aren't consistent between the two equations in your answer. $\endgroup$ Apr 19 '13 at 22:52
  • $\begingroup$ @Thomas: Ahem, Arjang did that when he edited the question. $\endgroup$ Apr 19 '13 at 22:53
  • $\begingroup$ @Zev : Do you think n should be used for upper limit and k as the index of summation? seeing k and usage like this is unfamiliar. Of course the answer is the same either way but it looks a bit odd $\endgroup$
    – jimjim
    Apr 19 '13 at 22:53
  • $\begingroup$ @Zev : sorry, was just writig to you about the edit. $\endgroup$
    – jimjim
    Apr 19 '13 at 22:54
  • $\begingroup$ @ThomasAndrews : My fault, apologies, I was confused by the usage of n and k, tried to edit to something more conventional ( I think). $\endgroup$
    – jimjim
    Apr 19 '13 at 22:56

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