evaluate the summation : $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(n+2x+3)}$ 
Find
$$S=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(n+2x+3)}$$
for $x≥0$.

At first, I use a partial fraction
$$S=\displaystyle\sum_{n=0}^{\infty}\left(\frac{(-1)^{n}}{2(x+1)(n+1)}-\frac{(-1)^{n}}{2(x+1)(n+2x+3)}\right)
=\frac{1}{2(x+1)}(I-J),$$
where
$$I=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1}
=\ln 2$$
and
$$J=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+3+2x}$$
I think to use:
$$\ln (1+y)=\sum_{n=0}^{\infty}\frac{(-1)^{n}y^{n+1}}{n+1}$$
Is my work correct? How to complete this work ?

We can write
$$
\eqalign{
  & S = \sum\limits_{n = 0}^\infty  {{{\left( { - 1} \right)^{\,n} } \over {\left( {n + 1} \right)\left( {n + 2x + 3} \right)}}}  =   \cr 
  &  = \sum\limits_{k = 0}^\infty  {\left( {{1 \over {\left( {2k + 1} \right)\left( {2k + 2x + 3} \right)}} - {1 \over {\left( {2k + 2} \right)\left( {2k + 2x + 4} \right)}}} \right)}  =
  \quad \quad (1) \cr 
  &  = {1 \over 4}\sum\nolimits_{k = 0}^{\,\infty } {{1 \over {\left( {k + 1/2} \right)\left( {k + x + 3/2} \right)}}}
  - {1 \over 4}\sum\nolimits_{k = 0}^{\,\infty } {{1 \over {\left( {k + 1} \right)\left( {k + x + 2} \right)}}}  =   \quad \quad (2) \cr 
  &  = {1 \over 4}\left( \matrix{
  \sum\nolimits_{k = 0}^{\,\infty } {{1 \over {\left( {k + 1/2} \right)\left( {k + 1/2 + x + 1} \right)}}}  +  \hfill \cr 
   - \sum\nolimits_{k = 0}^{\,\infty } {{1 \over {\left( {\left( {k + 1/2} \right) + 1/2} \right)\left( {\left( {k + 1/2} \right) + 1/2 + x + 1} \right)}}}  \hfill \cr}  \right) =   \cr 
  &  = {1 \over 4}\left( \matrix{
  \sum\nolimits_{k = 0}^{\,\infty } {{1 \over {\left( {k + 1/2} \right)\left( {k + 1/2 + x + 1} \right)}}}  +  \hfill \cr 
   - \sum\nolimits_{k = 1/2}^{\,\infty } {{1 \over {\left( {k + 1/2} \right)\left( {k + 1/2 + x + 1} \right)}}}  \hfill \cr}  \right) =   \cr 
  &  = {1 \over 4}\sum\nolimits_{k = 0}^{\,1/2} {{1 \over {\left( {k + 1/2} \right)\left( {k + 1/2 + x + 1} \right)}}}  =   \quad \quad (3) \cr 
  &  = {1 \over 4}\sum\nolimits_{k = 1/2\;}^{\,1} {{1 \over {k\left( {k + x + 1} \right)}}}  =  \quad \quad (4)  \cr 
  &  = {1 \over {4\left( {x + 1} \right)}}\sum\nolimits_{k = 1/2\;}^{\,1} {\left( {{1 \over k} - {1 \over {\left( {k + x + 1} \right)}}} \right)}  =  \quad \quad (5)  \cr 
  &  = {1 \over {4\left( {x + 1} \right)}}\left( {\psi \left( 1 \right) - \psi \left( {1/2} \right) - \left( {\psi \left( {x + 2} \right) - \psi \left( {x + 3/2} \right)} \right)} \right) =
  \quad \quad (6)  \cr 
  &  = {1 \over {4\left( {x + 1} \right)}}\left( {2\ln 2 - \psi \left( {x + 2} \right) + \psi \left( {x + 3/2} \right)} \right)    \quad \quad (7)   \cr
   } 
$$
where:
 (1) we separate even and odd $n$;
 (2) each part is convergent and we can separate into two sums,
at the same time we change the sum into [antidifference][1];
(3) (4)we change the indices and use the properties of anti-diffence;
(5) being the sum limits finite we can apply partial fractions ans separate the sums;
(6) the anti-difference of $1/k$ is the [digamma function][2];
(7) we use the known values for $\psi (1)$ and $\psi(1/2)$.
The final formula can be converted into many equivalent forms
and series developments, depending on the intended scopes.
