Sum of two harmonic alternating series 
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.
 A: 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).
$$\begin{eqnarray*}\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}(2n+1)}{n(n+1)}&=&\sum_{n=1}^{+\infty}(-1)^{n+1}\int_{0}^{1}(x^{n-1}+x^{n})\,dx\\&=&\int_{0}^{1}(1+x)\sum_{n\geq 1}(-1)^{n-1}x^{n-1}\,dx\\&=&\int_{0}^{1}\frac{1+x}{1+x}\,dx=\color{red}{1}.\end{eqnarray*}$$
A: $$
\sum_{n=1}^{\infty}{\frac{{(-1)}^{n+1}}{n+1}}=\sum_{n=0}^{\infty}{\frac{{(-1)}^{n+1}}{n+1}}+1\\
=\sum_{n=1}^{\infty}{\frac{{(-1)}^{n}}{n}}+1
$$
So summing the two summation yields:
$$
\sum_{n=1}^{\infty} {{(-1)}^{n+1}\frac{2n+1}{n(n+1)}}=\sum_{n=1}^{\infty}{\frac{{(-1)}^{n+1}}{n+1}}+\sum_{n=1}^{\infty}{\frac{{(-1)}^{n+1}}{n}}\\
=\sum_{n=1}^{\infty}{\frac{{(-1)}^{n}}{n}}+1-\sum_{n=1}^{\infty}{\frac{{(-1)}^{n}}{n}}=1
$$
A: \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}
A: \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.
