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.

\begin{align}
?
&\equiv
\sum\limits_{n = 1}^{\infty}\left(-1\right)^n \frac{2n + 3}{n\left(n + 1\right)}
=
\sum_{n = 1}^{\infty}\left\lbrack%
{4n + 3 \over 2n\left(2n + 1\right)}
-
{4n + 1 \over \left(2n - 1\right)\left(2n\right)}
\right\rbrack 
\\[3mm]&=
\sum_{n = 1}^{\infty}\left\lbrack%
{1 \over n}
+
{1 \over 2n\left(2n + 1\right)}
-
{1 \over n}
-
{3 \over \left(2n - 1\right)\left(2n\right)}
\right\rbrack
\\[3mm]&=
{1 \over 4}
\sum_{n = 1}^{\infty}\left\lbrack%
{1 \over n\left(n + 1/2\right)}
-
{3 \over n\left(n - 1/2\right)}
\right\rbrack
=
{1 \over 4}
\sum_{n = 0}^{\infty}\left\lbrack%
{1 \over \left(n + 1\right)\left(n + 3/2\right)}
-
{3 \over \left(n + 1\right)\left(n + 1/2\right)}
\right\rbrack
\\[3mm]&=
{1 \over 4}\left\lbrack%
{\Psi\left(3/2\right) - \Psi\left(1\right) \over 3/2 - 1}
-
3\,{\Psi\left(1\right) - \Psi\left(1/2\right) \over 1 - 1/2}
\right\rbrack
=
{1 \over 2}\left\lbrack%
\Psi\left(3 \over 2\right) - 4\Psi\left(1\right) + 3\Psi\left(1 \over 2\right) 
\right\rbrack
\\[3mm]&=
{1 \over 2}\left\lbrace%
\left\lbrack\Psi\left(1 \over 2\right) + 2\right\rbrack- 4\Psi\left(1\right) + 3\Psi\left(1 \over 2\right) 
\right\rbrace
=
1
+
2\left\lbrack\Psi\left(1 \over 2\right) - \Psi\left(1\right)\right\rbrack
\\[3mm]&=
1
+
2\left\lbrace%
\left\lbrack-\gamma - 2\ln\left(2\right)\right\rbrack
-
\left(-\gamma\right)
\right\rbrace
\end{align}
$$
\begin{array}{|c|}\hline\\
{\large%
?
\equiv
\sum\limits_{n = 1}^{\infty}\left(-1\right)^n \frac{2n + 3}{n\left(n + 1\right)}
=
\color{#ff0000}{1 - 4\ln\left(2\right)}}
\\ \\ \hline
\end{array}
$$
