infinite summation of exponential functions

I've discovered through Wolfram Alpha that

$\sum_{t=1}^{\infty}{e^{-bt}}=\frac{1}{e^b-1}$

What are the steps of derivation here? According to infinite summation of power series:

$\sum_{t=1}^{\infty}p^t=\frac{1}{1-p}$,

I expected the solution to be

$\sum_{t=1}^{\infty}{(e^{-b})^t}=\frac{1}{1-e^{-b}}$.

What am I getting wrong?

In extension, how do I derive

$\sum_{t=1}^{\infty}{e^{-b(t-1)}}$ ?

• For the last question, note that $e^{-b(t-1)}=e^be^{-bt}$, so just factor the $e^b$ out of the sum. – carmichael561 Aug 9 '16 at 18:45

Your second formula isn't quite right: if $|p|<1$, then $$\sum_{t=1}^{\infty}p^t=\frac{p}{1-p}$$ Using this with $p=e^{-b}$ yields $$\sum_{t=1}^{\infty}e^{-bt}=\frac{e^{-b}}{1-e^{-b}}=\frac{1}{e^b-1}$$ as claimed.

• perhaps the issues was one of starting at t=0 or t=1? – Kitter Catter Aug 9 '16 at 18:31
• Yes, that is of course the issue. – carmichael561 Aug 9 '16 at 18:32