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Evaluate the series $$\sum_{n=1}^\infty (-1)^{n+1}\frac{2n+1}{n(n+1)}.$$

I've simplified it to the form $$\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n+1} + \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}$$ and I've proved that both parts converge. However, I'm having trouble finding the limit. Writing out the terms as $(1+\frac 1 2 - \frac 1 2 + \frac 1 3 - \frac 1 3 ... )$ suggest their sum is one. However when I look up the sums of the two parts, they are $-\ln(2)$ and $\ln(2)$ respectively, which suggests the sum of the overall series is $0$. I'm aware that if a series is not absolutely convergent then its terms can be rearranged to converge to any number, but we haven't covered that topic yet so I feel like that shouldn't be a consideration in solving this.

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$$ \sum_{n=1}^{\infty}{\frac{{(-1)}^{n+1}}{n+1}}=\sum_{n=0}^{\infty}{\frac{{(-1)}^{n+1}}{n+1}}+1\\ =\sum_{n=1}^{\infty}{\frac{{(-1)}^{n}}{n}}+1 $$

So summing the two summation yields:

$$ \sum_{n=1}^{\infty} {{(-1)}^{n+1}\frac{2n+1}{n(n+1)}}=\sum_{n=1}^{\infty}{\frac{{(-1)}^{n+1}}{n+1}}+\sum_{n=1}^{\infty}{\frac{{(-1)}^{n+1}}{n}}\\ =\sum_{n=1}^{\infty}{\frac{{(-1)}^{n}}{n}}+1-\sum_{n=1}^{\infty}{\frac{{(-1)}^{n}}{n}}=1 $$

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You have to be careful in performing rearrangements since your series is conditionally convergent but not absolutely convergent (see, for instance, the Riemann series theorem).

$$\begin{eqnarray*}\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}(2n+1)}{n(n+1)}&=&\sum_{n=1}^{+\infty}(-1)^{n+1}\int_{0}^{1}(x^{n-1}+x^{n})\,dx\\&=&\int_{0}^{1}(1+x)\sum_{n\geq 1}(-1)^{n-1}x^{n-1}\,dx\\&=&\int_{0}^{1}\frac{1+x}{1+x}\,dx=\color{red}{1}.\end{eqnarray*}$$

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    $\begingroup$ I do not agree cancelling terms in telescopic series is really rearranging. It is rather rewriting the partial sums S_n as 1+1/n, so it is allowed. $\endgroup$ – Miguel Sep 4 '15 at 18:16
  • $\begingroup$ @MiguelAtencia: I was talking about simplifying the series into $\sum_{n=1}^{+\infty}-\sum_{n=1}^{+\infty}$. That clearly does not work for series like $\sum_{n=1}^{+\infty}\frac{(-1)^n}{n}$, but I agree with you that we may re-arrange partial sums in any way we like, till they are just partial sums. $\endgroup$ – Jack D'Aurizio Sep 5 '15 at 1:44
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\begin{align} \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \, (2n+1)}{n \, (n+1)} &= \sum_{n=1}^{\infty} (-1)^{n+1} \, \left( \frac{1}{n} + \frac{1}{n+1} \right) \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} + \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n} \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} + \left( \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} + 1 \right) \\ &= 1. \end{align}

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\begin{align} \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n+1} + \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}&= \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n+1} - \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}\\ &= \sum_{n=1}^\infty \Big((-1)^{n+1}\frac{1}{n+1} - (-1)^{n}\frac{1}{n}\Bigg)\\ &= \sum_{n=1}^\infty a_{n+1}-a_{n} \end{align} with $a_n=(-1)^{n}\frac{1}{n}$. Now use telescopic rule.

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