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.

$$
\sum_{n=1}^{\infty}{\frac{{(-1)}^{n+1}}{n+1}}=\sum_{n=0}^{\infty}{\frac{{(-1)}^{n+1}}{n+1}}+1\\
=\sum_{n=1}^{\infty}{\frac{{(-1)}^{n}}{n}}+1
$$
So summing the two summation yields:
$$
\sum_{n=1}^{\infty} {{(-1)}^{n+1}\frac{2n+1}{n(n+1)}}=\sum_{n=1}^{\infty}{\frac{{(-1)}^{n+1}}{n+1}}+\sum_{n=1}^{\infty}{\frac{{(-1)}^{n+1}}{n}}\\
=\sum_{n=1}^{\infty}{\frac{{(-1)}^{n}}{n}}+1-\sum_{n=1}^{\infty}{\frac{{(-1)}^{n}}{n}}=1
$$
