calculate the sum of (-1)^n (n/8^n)
i know fourier coefficients and bessels inequality how to evaluate this
The sum, $\displaystyle \sum_{n=1}^{\infty} (-1)^n\frac{n}{8^n}$, can be evaluated by using only elementary techniques.
Let $\displaystyle S=\sum_{n=1}^{\infty} (-1)^n\frac{n}{8^n}$...(1)
$\displaystyle 8S=\sum_{n=1}^{\infty} (-1)^n\frac{n}{8^{n-1}}$...(2)
Adding (1) and (2) gives the incredible result
$\displaystyle 9S=-1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{8^n}=-1+\frac{1}{9}$
$\displaystyle =\frac{-8}{9}$
so $\displaystyle S=\frac{-8}{81}$
$\displaystyle 8S=-1+\frac{2}{8}-\frac{3}{8^2}+\cdots$
$\displaystyle S=-\frac{1}{8}+\frac{2}{8^2}-\frac{3}{8^3}+\cdots$
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add them...
$\displaystyle 9S=-1+\frac{1}{8}-\frac{1}{8^2}+\frac{1}{8^3}-\cdots$
$\displaystyle =-1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{8^n}$
You posted this in the calculus forum. So I assume it should be done using calculus.
If $\displaystyle y = \sum\limits_{k = 1}^\infty {r^k } = \frac{r}{{1 - r}},\;\left| r \right| < 1$ then $\displaystyle y' = \sum\limits_{k = 1}^\infty {kr^{k - 1} } = \frac{1}{{\left( {1 - r} \right)^2 }}$.
The multiply by $\displaystyle r$ to get $\displaystyle \sum\limits_{k = 1}^\infty {kr^{k} } = \frac{r}{{\left( {1 - r} \right)^2 }}$.
In this question $\displaystyle r=\frac{-1}{8}$.