Compute the limit $$\sum_{n=1}^{\infty} \frac{n}{2^n}$$
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17I found it! Your turn! – Asaf Karagila Mar 12 '13 at 10:20
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3See this post and its linked pages for ideas. – David Mitra Mar 12 '13 at 10:37
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Me too, Asaf! Hint to OP: Do not use imperative when asking a question here. It's a violation of site etiquette. – ncmathsadist May 12 '15 at 01:31
3 Answers
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Hint: Consider the power series with those coefficients, for $x=1$. Remember that we can integrate term-by-term within the convergence radius.
Asaf Karagila
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3Yes, but I'm a set theorist. And I don't have to teach calculus this semester! So... jokes on someone else. – Asaf Karagila Mar 12 '13 at 10:22
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Hints:
(1) $f(x)=\frac1{1-x}=\sum_{n=0}^\infty x^n$ converges uniformly for $|x|\leq r$ for any $r<1$ and hence is differentiatable.
(2) $f'(x)=\sum_{n=1}^\infty nx^{n-1}=\frac1x\sum_{n=1}^\infty nx^{n}$
Dennis Gulko
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Hint: If you use
1.the software parses it as an ordered list (you may have to line break one more) – Asaf Karagila Mar 12 '13 at 10:22
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By ratio test, $\displaystyle \lim_{x\to \infty}\frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^{n}}}=\lim_{x\to \infty}\frac{n+1}{2n}=\frac 1 2<1$.
Therefore $\displaystyle\sum_{n=1}^\infty \frac n {2^n}$ convergent.
$\displaystyle S=\sum_{n=1}^\infty \frac n {2^n}=\frac 1 2 + \frac 2 {2^2} + \frac 3 {2^3}+...$ (i)
$2*$(i): $\displaystyle 2S=1 + \frac 2 {2} + \frac 3 {2^2}+...$ (ii)
(ii) $-$ (i): $\displaystyle S=1+\frac 1 2 + \frac 1 {2^2} +...$
Can you take it from here? (Answer: $S=2$)
p.s. This method requires just elementary skills. :)
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Maybe it should be added that this works if we know that the series converges (i.e., $S\ne+\infty$). – Martin Sleziak Mar 12 '13 at 13:30
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1@ Martin Sleziak: You are right, otherwise, $\infty-\infty$ is indeterminate. – JSCB Mar 12 '13 at 13:32
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If we subtract $S$ from $2S$, how can we conclude that the difference is equal to $S$ ? – J. Dunivin May 11 '15 at 18:57