The series is $1\cdot\frac{1}{2} + 2\cdot\frac{1}{4} + 3\cdot\frac{1}{8} + \cdots$
Or in other words
$$\sum_{n=1}^{\infty}\frac{n}{2^n}$$
What kind of series is this and how to find the sum? Thanks....
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2Hint. Differentiate the power series for $1/(1-x)$, then multiply by $x$. – Ethan Bolker Sep 26 '17 at 00:20
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That sounds like a great idea! Why don't you do it and I'll give you a lovely green check mark :) – Ben S. Sep 26 '17 at 00:25
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I think the third term should have $\frac18$. – marty cohen Sep 26 '17 at 00:29
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@martycohen fixed thanks. – Ben S. Sep 26 '17 at 00:30
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Nice accepted answer so I won't post mine. It's a good technique to remember. – Ethan Bolker Sep 26 '17 at 12:17
1 Answers
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Without calculus:
If $s(a) =\sum_{n=0}^{\infty} na^n $ for $|a| < 1$, then
$\begin{array}\\ as(a) &=\sum_{n=0}^{\infty} na^{n+1}\\ &=\sum_{n=1}^{\infty} (n-1)a^{n}\\ &=\sum_{n=1}^{\infty} na^{n}-\sum_{n=1}^{\infty} a^{n}\\ &=s(a)-\dfrac{a}{1-a}\\ \text{so}\\ s(a) &=\dfrac{a}{(1-a)^2}\\ \end{array} $
marty cohen
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Thank you. I am a little confused by one thing. You have $\sum_{n=1}^{\infty} a^{n} = a/(1-a)$ but I thought that it was $\sum_{n=0}^{\infty} a^{n}$ that summed to that? – Ben S. Sep 26 '17 at 00:40
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$\sum_{n=0}^{\infty} a^{n} = 1/(1-a)$. $\sum_{n=k}^{\infty} a^{n} = a^k/(1-a)$. – marty cohen Sep 26 '17 at 02:05
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