alternating series sum

**prasum** · November 23rd 2012, 10:55 PM

calculate the sum of (-1)^n (n/8^n)

i know fourier coefficients and bessels inequality how to evaluate this

**sbhatnagar** · November 24th 2012, 03:38 AM

The sum, $\sum_{n=1}^{\infty} (-1)^n\frac{n}{8^n}$ , can be evaluated by using only elementary techniques.

Let $S=\sum_{n=1}^{\infty} (-1)^n\frac{n}{8^n}$ ...(1)

$8S=\sum_{n=1}^{\infty} (-1)^n\frac{n}{8^{n-1}}$ ...(2)

Adding (1) and (2) gives the incredible result

$9S=-1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{8^n}=-1+\frac{1}{9}$
$=\frac{-8}{9}$

so $S=\frac{-8}{81}$

**prasum** · November 24th 2012, 04:02 AM

i did not understand the step which produces the incredible result can yu please explain

**sbhatnagar** · November 24th 2012, 04:45 AM

Originally Posted by prasum

i did not understand the step which produces the incredible result can yu please explain

$8S=-1+\frac{2}{8}-\frac{3}{8^2}+\cdots$

$S=-\frac{1}{8}+\frac{2}{8^2}-\frac{3}{8^3}+\cdots$
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add them...
$9S=-1+\frac{1}{8}-\frac{1}{8^2}+\frac{1}{8^3}-\cdots$

$=-1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{8^n}$

**Plato** · November 24th 2012, 04:49 AM

Originally Posted by prasum

i did not understand the step which produces the incredible result can yu please explain

You posted this in the calculus forum. So I assume it should be done using calculus.

If $y = \sum\limits_{k = 1}^\infty {r^k } = \frac{r}{{1 - r}},\;\left| r \right| < 1$ then $y' = \sum\limits_{k = 1}^\infty {kr^{k - 1} } = \frac{1}{{\left( {1 - r} \right)^2 }}$ .

The multiply by $r$ to get $\sum\limits_{k = 1}^\infty {kr^{k} } = \frac{r}{{\left( {1 - r} \right)^2 }}$ .

In this question $r=\frac{-1}{8}$ .

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