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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!

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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 $N \rightarrow \infty$

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  • $\begingroup$ Oh I see, so the 2nd one is like a telescoping series kind of thing! My teacher said the answer is 1, does (−1)^N/(N+2) cancel out somehow? Thanks for your help! $\endgroup$ – Matthew May 10 '11 at 5:44
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    $\begingroup$ Cancel out? No. Approach zero as $N$ increases? Yes. $\endgroup$ – Gerry Myerson May 10 '11 at 5:49
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    $\begingroup$ @user4773: the term $(-1)^N/(N+2)$ goes to 0 as $N\to\infty$. $\endgroup$ – user10287 May 10 '11 at 5:50
  • $\begingroup$ Ah, thanks. I'm always forgetting to take limits. $\endgroup$ – Matthew May 10 '11 at 5:53
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$\begin{aligned} \displaystyle \sum_{n\ge 0}\left(\frac{(-1)^n}{n+1}+\frac{(-1)^n}{(n+2)}\right) & = \sum_{n\ge 0}\left(\int_{0}^{1}(-1)^nx^{n}\;{dx}+\int_{0}^{1}(-1)^nx^{n+1}\;{dx}\right) \\& = \int_{0}^{1}\left(\sum_{n\ge 0}(-1)^nx^{n}+\sum_{n\ge 0}(-1)^nx^{n+1}\right)\;{dx} \\& = \int_{0}^{1} \frac{1}{1+x}+\frac{x}{1+x}\;{dx} \\& \\& = \int_{0}^{1} \;{dx} \\& = 1. \end{aligned}$

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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: $$ \sum_{n=1}^\infty nx^n = \sum_{m \geq 1} \sum_{n \geq m} x^n = \sum_{m \geq 1} \frac{x^m}{1-x} = \frac{x}{(1-x)^2}. $$

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  • $\begingroup$ Thanks for your help, I did write out the first few terms but summed up the two parts of the series instead of keeping them separate so it wasn't as clear as it should have been. $\endgroup$ – Matthew May 10 '11 at 5:44
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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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