Finding the sum of a series
I've been having some problems with these two series.
$$\begin{align}&\sum_{n=0}^\infty (-1)^n\frac{n}{3^n}\\&\sum_{n=0}^\infty \left(\frac{(-1)^n}{n+1}+\frac{(-1)^n}{n+2}\right)\end{align}$$
The first seems similar to a geometric series but I don't know what to do with that extra $n$.
The second I really just don't know what to do either.
I'd really appreciate any help you can give me, thanks!
$\endgroup$4 Answers
$\begingroup$For the first sum, consider $f(x) = \displaystyle \sum_{n=0}^{\infty} x^n$ where $|x| < 1$. We have that $f(x) = \frac1{1-x}$ (geometric series)
$f'(x) = \displaystyle \sum_{n=1}^{\infty} n x^{n-1} = \frac1{(1-x)^2}$. Now plug in $x = -\frac1{3}$ to get what you want.
For the second one, look at the partial sums i.e. let $S_N = \displaystyle \sum_{n=0}^{N} \left( \frac{(-1)^n}{n+1} + \frac{(-1)^n}{n+2} \right) = \left( \frac1{1} + \frac1{2} - \frac1{2} - \frac1{3} + \cdots + \frac{(-1)^N}{N+1} + \frac{(-1)^N}{N+2} \right)$
Hence, $S_N = \left( 1 + \frac{(-1)^N}{N+2}\right)$. Now take the limit as $N \rightarrow \infty$
$\endgroup$ 4 $\begingroup$$\begin{aligned} \displaystyle \sum_{n\ge 0}\left(\frac{(-1)^n}{n+1}+\frac{(-1)^n}{(n+2)}\right) & = \sum_{n\ge 0}\left(\int_{0}^{1}(-1)^nx^{n}\;{dx}+\int_{0}^{1}(-1)^nx^{n+1}\;{dx}\right) \\& = \int_{0}^{1}\left(\sum_{n\ge 0}(-1)^nx^{n}+\sum_{n\ge 0}(-1)^nx^{n+1}\right)\;{dx} \\& = \int_{0}^{1} \frac{1}{1+x}+\frac{x}{1+x}\;{dx} \\& \\& = \int_{0}^{1} \;{dx} \\& = 1. \end{aligned}$
$\endgroup$ $\begingroup$Hint for the first series: Expand it as a sum of geometric series. This is the most straightforward way to solve this, though there are others.
Hint for the second series: Write the first few terms and see what you get.
Edit: Here's how the first hint could be used: $$ \sum_{n=1}^\infty nx^n = \sum_{m \geq 1} \sum_{n \geq m} x^n = \sum_{m \geq 1} \frac{x^m}{1-x} = \frac{x}{(1-x)^2}. $$
$\endgroup$ 1 $\begingroup$For the first one, let $s_{n} = \sum_{k=1}^{n} k x^{k}$. Then, $$ \begin{aligned} xs_{n}&=\sum_{k=1}^{n}kx^{k+1}\\ &=\sum_{k=1}^{n}(k+1-1)x^{k+1}\\ &=\sum_{k=2}^{n+1}(k-1)x^{k}\\ &=nx^{n+1}+\sum_{k=1}^{n}(k-1)x^{k}\\ &=nx^{n+1}+s_{n} - \sum_{k=1}^{n}x^{k}\\ &=nx^{n+1}+s_{n} -\frac{x - x^{n+1}}{1-x}. \end{aligned} $$ Now, solve for $s_{n}$, set $x = -1/3$ and let $n\to\infty$.
For the second one, note that $$ \frac{(-1)^n}{n+1} + \frac{(-1)^n}{n+2} = \frac{(-1)^n}{n+1}-\frac{(-1)^{n+1}}{n+2}. $$
Let $w_{n} = (-1)^n/(n+1)$. Then, the second series reduces to $$\begin{aligned} \lim\sum_{n=0}^{m} (w_{n}-w_{n+1})&=\lim\left[-\sum_{n=0}^{m}(w_{n+1}-w_{n})\right]\\&=\lim(w_{0}-w_{m+1})\\&=\lim\left(1+\frac{(-1)^m}{m+2}\right)\\&=1. \end{aligned}$$
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