This is a homework question; I'm supposed to use power series to find the following sum: $$\sum_{n=1}^\infty \frac{n}{2^n}$$ I took the geometric series $$\frac{1}{1-x}=\sum_{n=0}^\infty {x^n}$$ and differentiated and multiplied both sides by x to get $$\frac{x}{(1-x)^2}=\sum_{n=1}^\infty {nx^n}$$ I'm stuck because I'm not sure how to make the $$\frac{1}{2^n}$$ term appear.
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7Try $x=\frac 12$? – Mark Bennet Mar 19 '14 at 08:43
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1http://en.wikipedia.org/wiki/Geometric_series#Geometric_power_series – lab bhattacharjee Mar 19 '14 at 08:44
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As an aside, this type of infinite series is called a polylogarithm. – Lucian Mar 19 '14 at 09:17
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Hem, what's the connection between $x^n$ and $1/2^n$ ?? – Mar 28 '17 at 10:03
3 Answers
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For $x=\frac {1} {2}$, we have $\sum _{i} nx^n=\sum _{i} n(\frac {1} {2})^n=\sum _{i} \frac {n} {2^n}$
Astro Nauft
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Without power series:
$$2S-S:=\frac11+\frac22+\frac34+\frac48\cdots-\frac12-\frac24-\frac38-\cdots =\frac11+\frac12+\frac14+\frac1{8}+\cdots=2.$$
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You are basically all of the way there. In your last step, you have $$ \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^n, \tag{$\ast$}$$ where the series on the right converges for any $|x| < 1$ to the rational expression on the left. Since we get to pick $x$ to be anything we like (as long as $|x| < 1$), we might as well choose $x = \frac{1}{2}$. If we do this, the series on the right-hand side of ($\ast$) becomes $$ \sum_{n=1}^{\infty} n \left( \frac{1}{2} \right)^n = \sum_{n=1}^{\infty} n \cdot \frac{1}{2^n} = \sum_{n=1}^{\infty} \frac{n}{2^n}, $$ which is the original series that you were given to evaluate. But with $x = \frac{1}{2}$, the left-hand side of ($\ast$) is $$ \frac{1}{\left( 1 - \frac{1}{2} \right)^2} = \frac{1}{\left(\frac{1}{2}\right)^2} = \frac{1}{\frac{1}{4}} = 4.$$ Therefore, equating the left- and right-hand sides of ($\ast$) when $x = \frac{1}{2}$, we obtain $$ \sum_{n=1}^{\infty} \frac{n}{2^n} = 4.$$
Xander Henderson
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