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

$$ \sum_{k=1}^{n}\frac{(-1)^{k+1}}{k(k+1)}=\sum_{k=1}^{n}(-1)^{k+1}\left(\frac1k-\frac1{k+1}\right) =\sum_{k=1}^n\frac{(-1)^{k+1}}{k}+\sum_{k=1}^n\frac{(-1)^{k+2}}{k+1} \\=2\sum_{k=1}^n\frac{(-1)^{k+1}}{k}-1-\frac{(-1)^{n+1}}{n+1}. $$ Hence, $$ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}=2\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}-1=2\ln 2-1, $$ since $$ \sum_{n=1}^n\frac{(-1)^{n+1}}{n}=\ln 2. $$

Proof. If $\lvert x\rvert<1$, then $$ \frac{1}{1+x}=\sum_{k=0}^n(-1)^{k}x^k+\frac{(-1)^{n+1}x^{n+1}}{1+x}, $$ and hence $$ \log(1+x)=\int_0^x\frac{dt}{1+t}=\sum_{k=0}^n(-1)^{k}\int_0^tt^k\,dt+\int_0^x\frac{(-1)^{n+1}t^{n+1}\,dt}{1+t} \\=\sum_{k=0}^n\frac{(-1)^kx^{n+1}}{n+1}+R_n(x). $$ Clearly, for $x\in [0,1]$ $$ \lvert R_n(x)\rvert = \int_0^x\frac{t^{n+1}\,dt}{1+t}\le\int_0^xt^{n+1}\,dt\le \frac{1}{n+2}. $$ Hence $$ \ln 2=\lim_{x\to 1^-}\ln(1+x)=\lim_{x\to 1^-}\sum_{k=0}^n\frac{(-1)^kx^{n+1}}{n+1}+\lim_{x\to 1^-}R_n(x)=\sum_{k=0}^n\frac{(-1)^k}{n+1}+R_n(1), $$ and hence $$ \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}=\lim_{n\to\infty}\sum_{k=0}^n\frac{(-1)^k}{n+1}=\lim_{n\to\infty}\big(\ln 2+R_n(1)\big)=\ln 2. $$

Using falling factorials and finite calculus we can write

$$\begin{align}\sum\frac{(-1)^{n+1}}{n(n+1)}\delta n&=\sum (-1)^{n+1}(n-1)^{\underline{-2}}\delta n\\&=(-1)^{n}(n-1)^{\underline{-1}}-\sum\frac{(-1)^{n+2}-(-1)^{n+1}}{(-1)(n+1)}\delta n\\&=\frac{(-1)^n}n+2\sum\frac{(-1)^{n}}{n+1}\delta n\end{align}$$

where the second step is summation by parts (check this, by example), and the indefinite sum of a falling factorial is defined as

$$\sum n^{\underline m}\delta n=\frac{n^{\underline{m+1}}}{m+1}+ K$$

Now taking limits above we have

$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}=\sum\nolimits_1^\infty\frac{(-1)^{n+1}}{n(n+1)}\delta n=\frac{(-1)^n}n\bigg|_1^\infty+2\sum\nolimits_1^\infty\frac{(-1)^{n}}{n+1}\delta n=\\=1+2(\ln (2)-1)=2\ln(2)-1$$

because the Taylor expansion of $\ln (1+x)$ is

$$\ln (1+x)=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^k}{k},\quad\text{whenever }x\in(-1,1]$$

I got my answer in this way: $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}=\sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac1n-\frac1{n+1}\right)=\sum_{n=1}^{\infty}\frac {(-1)^{n+1}}{n}-\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n+1}=\\log2+\log2-1=2\log2-1$