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What is the sum of the series $\sum_{n=0}^\infty\left(\frac{4n+3}{5^n}\right)$ ? I got that the series converges and the sum seems to be $5$. When trying to explicitly get the sum, I tried to find the appropriate function whose Maclaurin series would be this sum.

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  • $\begingroup$ $\sum_{n=1}^N\left(\frac{4n+3}{5^n}\right)=2-\frac{2+N}{5^N}$ $\endgroup$ – d.k.o. Jun 21 '15 at 1:29
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$$\sum_{n=0}^{\infty }\frac{4n+3}{5^n}=\sum_{n=0}^{\infty }\frac{4n}{5^n}+\sum_{n=0}^{\infty }\frac{3}{5^n}$$ $$\sum_{n=0}^{\infty }x^n=\frac{1}{1-x}$$ derive it $$\sum_{n=1}^{\infty }nx^{n-1}=\frac{1}{(1-x)^2}$$ multiply by $x$ $$\sum_{n=1}^{\infty }nx^{n}=\frac{x}{(1-x)^2}$$ at $x=1/5$ $$\sum_{n=1}^{\infty }\frac{4n}{5^{n}}=4\frac{1/5}{(1-1/5)^2}$$ the second series is geometric series $$3\sum_{n=0}^{\infty }x^n=3\left(\frac{1}{1-x}\right)$$ find the value at $x=1/5$

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    $\begingroup$ You really don't need to subtract one before taking the derivative. $\endgroup$ – Thomas Andrews Jun 21 '15 at 1:39
  • $\begingroup$ @ThomasAndrews but I think the $n$ was $1$ $\endgroup$ – user249369 Jun 21 '15 at 1:45
  • $\begingroup$ Yes, but what is $nx^n$ when $n=0$? $\endgroup$ – Thomas Andrews Jun 21 '15 at 1:46
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Hint: $$\sum_{n=0}^\infty \frac{n}{5^n}=\sum_{n=1}^\infty\sum_{m=n}^\infty \frac{1}{5^m}$$

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a telescoping way,

$\sum_{n=0}^\infty\left(\frac{n+1}{5^{n-1}}\right)$$-$ $\sum_{n=0}^\infty\left(\frac{n+2}{5^n}\right)$

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  • $\begingroup$ I was thinking in this line. But we need to prove the convergence, right? Also, how to derive this relation? $\endgroup$ – lab bhattacharjee Jun 21 '15 at 2:52
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$$ \begin{align} \sum_{n=0}^{\infty}(4n+3)x^n&=4\sum_{n=0}^{\infty}(n+1)x^n\sum_{n=0}^{\infty}x^n\\ &=4(\sum_{n=0}^{\infty}x^{n+1})'-\sum_{n=0}^{\infty}x^n\\ &=4(\frac{x}{1-x})'-\frac{1}{1-x}\\ &=\frac{x+3}{(1-x)^2} \end{align} $$

Set $x=\frac{1}{5}$, we have $\sum_{n=0}^{\infty}\frac{4n+3}{5^n}=5$.

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