evaluate the summation : $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(n+2x+3)}$

Question

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 ?

Answer 1

$$S(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)(n+2x+3)}=\sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} \int _{0}^{1} t^{n+2x+2} dt=\int_{0}^{1} t^{2x+1} \ln(1+t) dt ~~~(1).$$ $$\Rightarrow S(x)=\left .\ln(1+t) \frac{t^{2x+2}}{2x+2}\right |_{0}^{1}-\frac{1}{2x+2}\int_{0}^{1} \frac{t^{2x+2}dt}{1+t}=\frac{\ln 2}{2x+2} - \frac{\beta(2x+3)} {2x+2}~~~~(2).$$ 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 $$ S(0)=0.25, S(1)=0.1458..=7/48, S(2)=0.1027=37/360, S(3)=0.7931=533/6720.$$

Answer 2

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 $A250344$.

Answer 3

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.