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.

\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}
