# Thread: series summation problem

1. ## series summation problem

$\displaystyle \sum _{n=0}^{\infty } \left(\frac{1}{2 (1+n) (1+2 n)}\right)=\text{Log}[2]$

2. ## Re: series summation problem

Originally Posted by capea
$\displaystyle \sum _{n=0}^{\infty } \left(\frac{1}{2 (1+n) (1+2 n)}\right)=\text{Log}[2]$
From the series expansion...

$\displaystyle \sum_{n=0}^{\infty} x^{2n}= \frac{1}{1-x^{2}}$ (1)

... we derive first...

$\displaystyle \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1} = \int_{0}^{x} \frac{d \xi}{1-\xi^{2}} = \frac{1}{2}\ \ln \frac{1+x}{1-x}$ (2)

... and then...

$\displaystyle \sum_{n=0}^{\infty} \frac{x^{2n+2}}{(2n+1)\ (2n +2)}= \frac{1}{2}\ \int_{0}^{x} \ln \frac{1+\xi}{1-\xi}\ d \xi =$

$\displaystyle = \frac{1}{2}\ \{(1+x)\ \ln (1+x) + (1-x)\ \ln (1-x) \}$ (3)

Now, setting in (3) $\displaystyle x=1$ You obtain...

$\displaystyle \sum_{n=0}^{\infty} \frac{1}{(2n+1)\ (2n +2)}= \ln 2$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

thanks

4. ## Re: series summation problem

Originally Posted by capea
$\displaystyle \sum _{n=0}^{\infty } \left(\frac{1}{2 (1+n) (1+2 n)}\right)=\text{Log}[2]$

$\displaystyle \frac{1}{2 (1+n) (1+2 n)}=\frac{1}{2n+1}-\frac{1}{2(n+1)}$

$\displaystyle \sum {\frac{1}{2n+1}-\frac{1}{2(n+1)}}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{(-1)^{k+1}}{k}+...=\sum {\frac{(-1)^{n+1}}{n}}=\ln2$