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December 5th, 2018, 06:51 AM   #1
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Series Computation

What is the general method to compute an infinite sum ?
I will compute one sum function of $\displaystyle f(n,p)=n(p+1)^{-n}$ and see if it works
Let $\displaystyle s(n)=\sum^{\infty}_{n=1} np^{-n}\; \; $ for $\displaystyle n \in N$ and $\displaystyle p>1$
$\displaystyle \sum^{\infty}_{n=1} x^n =\frac{1}{1-x} \; \; \; $ where $\displaystyle 0<x<1$
$\displaystyle (\sum^{\infty}_{n=1} x^n)' =(\frac{1}{1-x})' \; \Rightarrow \;
\sum^{\infty}_{n=1} (x^n )'=\frac{1}{(1-x)^2 }$
$\displaystyle \sum^{\infty}_{n=1}nx^{n-1} =\frac{1}{(1-x)^2 } \; \; $ or $\displaystyle \; \sum^{\infty}_{n=1}nx^{n}=\frac{x}{(1-x)^2 }$
for $\displaystyle x=p^{-1} \; \; $ then $\displaystyle s(n)=p (p-1)^{-2}$
$\displaystyle \sum^{\infty}_{n=1} np^{-n}=p(p-1)^{-2}$
Let's see examples for some values of $\displaystyle p$ below :
p=e then $\displaystyle \sum^{\infty}_{n=1} ne^{-n}=e(e-1)^{-2}$
p=3 then $\displaystyle \sum^{\infty}_{n=1} n3^{-n} = \frac{3}{4}$
p=2 then $\displaystyle \sum^{\infty}_{n=1} n2^{-n}=2$
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December 5th, 2018, 12:51 PM   #2
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As far as I know, "general" methods for this cover various special cases only.
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