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 ?
 A: If, as Dr Zafar Ahmed DSc answered, we use $$S_x=\int_{0}^{1} t^{2x+1} \log(1+t)\, dt$$ we should have
$$S_x=\frac{H_{x-\frac{1}{2}}-H_x}{4( x+1)}+\frac{1+2\log(2)+6\log(2)x +4\log(2)x^2} {4 (x+1)^2\, (2 x+1) }$$ which, for $0 \leq x \leq 9$, would generate the sequence
$$\left\{\frac{1}{4},\frac{7}{48},\frac{37}{360},\frac{533}{6720},\frac{1627}{25200},
   \frac{18107}{332640},\frac{237371}{5045040},\frac{95549}{2306304},\frac{1632341}
   {44108064},\frac{155685007}{4655851200}\right\}$$
Notice that the numerators correspond to sequence $A082687$ and, if divided by $4$, the denominators are given in $A250344$.
A: $$S(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)(n+2x+3)}=\sum_{n=0}^{\infty}
\frac{(-1)^n}{n+1} \int _{0}^{1} t^{n+2x+2} dt=\int_{0}^{1} t^{2x+1} \ln(1+t) dt ~~~(1).$$ $$\Rightarrow S(x)=\left .\ln(1+t) \frac{t^{2x+2}}{2x+2}\right |_{0}^{1}-\frac{1}{2x+2}\int_{0}^{1} \frac{t^{2x+2}dt}{1+t}=\frac{\ln 2}{2x+2} - \frac{\beta(2x+3)}
{2x+2}~~~~(2).$$
Here we have used 3.222 of Garadshteyn and Ryzhik where $$\int_{0}^{1}\frac{x^{{\mu}-1}}{1+x} dx= \beta(\mu).$$
Next $$\beta(x)=\frac{1}{2}[\psi(\frac{x+1}{2})-\psi(\frac{x}{2})]~~~~(3)$$
and $\psi(x)$ is Digamma function definded as $\psi(x)=\frac{\Gamma'(x)}
{\Gamma(x)}= -C -\sum \left( \frac{1}{x+k}-\frac{1}{k+1} \right),$ $C$ is Eular constant equa to =$-\psi(1)=0.5772....$
Using (3), we find that $$ S(0)=0.25, S(1)=0.1458..=7/48, S(2)=0.1027=37/360, S(3)=0.7931=533/6720.$$ 
A: 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.
