Sum $ \sum\limits_{k=1}^{n} (-1)^k \frac{2k+3}{k(k+1)} $ I have the following Sum
$$ \sum\limits_{n=1}^{\infty} (-1)^n \frac{2n+3}{n(n+1)}  $$
and I need to calculate the sums value by creating the partial sums.
I started by checking if  $$\sum\limits_{n=1}^{\infty} \left| (-1)^n \frac{2n+3}{n(n+1)}  \right|$$ converges.
i tried to check for convergence with the 2 criterias a) decreasing and b) zero sequence but yes then i tried to transform the equotation to
$$ \sum\limits_{n=1}^{\infty} \frac{2}{n+1} + \frac{1}{n(n+1)} $$
the first fraction is "ok" - the 2nd one i did partial fraction decomposition and finally got $$ \sum\limits_{n=1}^{\infty} \frac{2}{n+1} + \frac{1}{n} - \frac{1}{n+1} $$
i then tried to see a pattern by find out the first sequences but im not sure if i'm on the right track.

$\dfrac{2n+3}{n(n+1)} = \color{red}{\dfrac{3}{n}} - \color{blue}{\dfrac{1}{n+1}}$,
so
$$
\sum_{n=1}^{\infty} (-1)^n \dfrac{2n+3}{n(n+1)} 
=\color{red}{\sum_{n=1}^{\infty} (-1)^n \dfrac{3}{n}} - \color{blue}{\sum_{n=1}^{\infty} (-1)^n \dfrac{1}{n+1}}.
$$
(all series are convergent here).
A) 
$$
\color{red}{\sum_{n=1}^{\infty} (-1)^n \dfrac{3}{n}} = 3 \sum_{n=1}^{\infty}  \dfrac{(-1)^n}{n} = -3\ln(2), \tag{A}
$$
(see Mercator series, Taylor series);
B) 
$$
\color{blue}{\sum_{n=1}^{\infty} (-1)^n \dfrac{1}{n+1}} = \sum_{k=2}^{\infty}  \dfrac{(-1)^{k-1}}{k} = -1+\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k} = -1+\ln(2).  \tag{B}
$$
Applying $(A)-(B)$, we get:
$$
\sum_{n=1}^{\infty} (-1)^n \dfrac{2n+3}{n(n+1)} = -3 \ln(2) - (-1+\ln(2)) = 1-4\ln(2).
$$
