What is the value of the series $$\sum_{n=1}^{\infty} \frac{n}{2^{n}}?$$ It is neither geometric nor a telescopic series. How do we approach such kind of series, given that we know that they converge?
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1see: https://math.stackexchange.com/questions/2012468/prove-limit-of-sum-n-1-infty-n-2n – farruhota Jul 06 '17 at 12:09
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2Possible duplicate of Evaluate $\sum_{n=1}^\infty \frac{n}{2^n}$., or Compute the limit $\sum_{n=1}^{\infty} \frac{n}{2^n}$. – Frenzy Li Jul 06 '17 at 12:09
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@Prashant Mishra What these existing posts use, in essence, is the ratio test. This is one of 3 main test to examine the convergence of infinite series. – Toby Mak Jul 06 '17 at 12:10
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There are two approaches I know of:
$$\sum_{n=1}^\infty n (1/2)^n = \sum_{n=1}^\infty \sum_{m=1}^n (1/2)^n = \sum_{m=1}^\infty \sum_{n=m}^\infty (1/2)^n$$
and then use the general geometric series formula. This amounts to recognizing the sum as a sum over a triangular array summed in one order and then summing over that same array in the other order.
The other one: for $0<|r|<1$:
$$\frac{d}{dr} \sum_{n=1}^\infty r^n = \sum_{n=1}^\infty n r^{n-1} = \frac{1}{r} \sum_{n=1}^\infty n r^n.$$
So $\sum_{n=1}^\infty n r^n = r \frac{d}{dr} \sum_{n=1}^\infty r^n$. On the other hand that derivative is also $\frac{d}{dr} \frac{r}{1-r}$ again by the geometric series formula. So computing that derivative for general $r$ and plugging in $r=1/2$ gives the desired result.
Both of these approaches can also be modified to provide a formula for $\sum_{n=1}^N n (1/2)^n$. This is interesting in and of itself, but also one can then take the limit of them as $N \to \infty$. In the process one can avoid using "heavy" theorems about interchange of summation and summation or interchange of summation and differentiation.
Ian
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