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