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.
 A: $\dfrac{2n+3}{n(n+1)} = \color{red}{\dfrac{3}{n}} - \color{blue}{\dfrac{1}{n+1}}$,
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:
$$
\sum_{n=1}^{\infty} (-1)^n \dfrac{2n+3}{n(n+1)} = -3 \ln(2) - (-1+\ln(2)) = 1-4\ln(2).
$$
A: \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}
$$
\begin{array}{|c|}\hline\\
{\large%
?
\equiv
\sum\limits_{n = 1}^{\infty}\left(-1\right)^n \frac{2n + 3}{n\left(n + 1\right)}
=
\color{#ff0000}{1 - 4\ln\left(2\right)}}
\\ \\ \hline
\end{array}
$$
A: What you wrote, coupled with the fact that:
$$\sum_{n=1}^\infty\frac{(-1)^n}{n}=-\log 2$$
Should give you the result: $(1 - 4 \log2)$. You can use the Alternating series test to show that since tha absolute value of your series monotonically decreases and moreover goes to zero as $n\to\infty$, the series converges.
A: i think this is an easy answer
$$ \sum_{n=1}^{\infty} (-1)^n \frac{2n+3}{n(n+1)} $$ 
$$ \frac{2n+3}{n(n+1)} = \frac{3(n+1) - n}{n(n+1)} = \frac{3}{n} - \frac{1}{n+1} $$
$$ \sum_{n=1}^{\infty} \left(\frac{3(-1)^n}{n} - \frac{(-1)^n}{n+1} \right) $$
$$ = \int_0^1 3\sum_{n=1}^{\infty} (-1)^n x^{n-1} - \sum_{n=1}^{\infty} (-1)^n x^n \ dx $$
$$ = \int_0^1 -\frac{3}{1+x} + \frac{x}{x+1}  \ dx $$
$$ =\int_0^1 \frac{x+1 - 4}{x+1} \ dx = \int_0^1 1 - \frac{4
}{x+1} \ dx = \left[x - 4\ln(x+1) \right]_0^1 = 1 - 4\ln 2 $$
i think you know $$ \sum_{n=1}^{\infty} x^n = \frac{x}{1-x} $$
