I've discovered through Wolfram Alpha that


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


I expected the solution to be


What am I getting wrong?

In extension, how do I derive

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

  • 1
    $\begingroup$ For the last question, note that $e^{-b(t-1)}=e^be^{-bt}$, so just factor the $e^b$ out of the sum. $\endgroup$ – 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.

  • 1
    $\begingroup$ perhaps the issues was one of starting at t=0 or t=1? $\endgroup$ – Kitter Catter Aug 9 '16 at 18:31
  • 2
    $\begingroup$ Yes, that is of course the issue. $\endgroup$ – carmichael561 Aug 9 '16 at 18:32

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