$\sum _{x=1}^{\mathrm{\infty}}{e}^{-sx}p{q}^{x-1}=\frac{p{e}^{-s}}{1-q{e}^{-s}}$

Dulce Cantrell
2022-09-14
Answered

How the answer was computed for the following infinite series:

$\sum _{x=1}^{\mathrm{\infty}}{e}^{-sx}p{q}^{x-1}=\frac{p{e}^{-s}}{1-q{e}^{-s}}$

$\sum _{x=1}^{\mathrm{\infty}}{e}^{-sx}p{q}^{x-1}=\frac{p{e}^{-s}}{1-q{e}^{-s}}$

You can still ask an expert for help

kappastud98u

Answered 2022-09-15
Author has **10** answers

Hint:

${e}^{-sx}p{q}^{x-1}=({e}^{-s}q{)}^{x-1}\cdot p{e}^{-s}$

${e}^{-sx}p{q}^{x-1}=({e}^{-s}q{)}^{x-1}\cdot p{e}^{-s}$

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