# Is there an identity for $\sum\limits_{n=1}^\infty \frac{1}{k^n}$?

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?

-
not a good edit, who is m? – user67133 Apr 19 '13 at 21:46
@user67133 Fixed, thanks :) – KevinOrr Apr 19 '13 at 21:47
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. – Arjang Apr 19 '13 at 22:50
@Arjang: since there is an answer already using the original notation, let's leave it. – robjohn Apr 19 '13 at 23:13
@KevinOrr : Google geometric series – Stefan Smith 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 $|k|>1$.