$$\sum_{n=1}^{\infty} \frac{n}{4^{n+1}}$$
Can someone help me find the sum of the series? I keep getting the sum to be $ {\frac{1}{12}}$ but the answer is $ {\frac{1}{9}}$.
Very detailed explanation will be greatly appreciated.
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This is the case $x=1/4$ of the series $\sum_{n=1}^\infty nx^{n+1}$. – Angina Seng Apr 29 '18 at 02:07
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1Isn't just $\dfrac{1}{4^2}+\dfrac{2}{4^3}+\dfrac{3}{4^4}=\dfrac{27}{256}>\dfrac{1}{12}$? – user061703 Apr 29 '18 at 02:08
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how did you find $1/12$? – operatorerror Apr 29 '18 at 02:17
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Possible dulplicate of https://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-inftyn1xn – Tony Ma Apr 29 '18 at 02:24
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Remember that
$$ \frac{1}{1-x} = \sum_{n=0}^{+\infty} x^n $$
Take the derivative w.r.t $x$
$$ \frac{1}{(1-x)^2} = \sum_{n=1}^{+\infty}n x^{n-1} $$
Now use $x = 1/4$
$$ \frac{1}{(1 - 1/4)^2} = \sum_{n=1}^{+\infty}\frac{n}{4^{n-1}} = \sum_{n=0}^{+\infty}\frac{n}{4^{n-1}} $$
Can you take it from here?
caverac
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Recall that $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$ for $|x|<1$. So for us, the series is: $$\frac{1}{16}\sum_{n=1}^\infty \frac{n}{4^{n-1}}=\frac{1}{16}\frac{d}{dx}\left.\left(\sum_{n=0}^\infty x^n\right)\right\vert_{x=1/4}=\frac{1}{16}\left(\frac{1}{(1-1/4)^2}\right)$$
Dave
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