I was asked to sum the given series:
- $\displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n\cdot (2n-1)}=\frac{1}{1} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 5} + \cdots \infty$
Here i workout the details.
\begin{align*} \sum\limits_{n=1}^{\infty} \frac{1}{n \cdot (2n-1)} &= \frac{1}{1} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 5} + \cdots \infty \\ &= 1 + \Bigl( \frac{1}{2} - \frac{1}{3}\Bigr) + \frac{1}{2} \Bigl(\frac{1}{3} - \frac{1}{5}\Bigr) + \frac{1}{3} \Bigl(\frac{1}{4}-\frac{1}{7}\Bigr) + \cdots \infty \\ &= 1 + \biggl[ \frac{1}{1 \cdot 2 } + \frac{1}{2 \cdot 3} + \cdots \infty\biggr] - \biggl[ \frac{1}{1 \cdot 3} + \frac{1}{2 \cdot 5} + \frac{1}{3 \cdot 7} + \cdots \infty \biggr] \\ &= 2 - \sum\limits_{n=1}^{\infty} \frac{1}{n \cdot (2n+1)} \end{align*}
Now for summing that second sum note that $-\log(1-x^{2}) =-\biggl[x^{2} + \frac{x^{4}}{2} + \frac{x^{6}}{3} + \cdots \infty\biggr]$
Have i done this correctly or is there some error in computing the sum. Now what i need to evaluate is the integral $\int\limits_{0}^{1} \log(1-x^{2}) \ \rm{dx}$
Can anyone tell me how to evaluate this integral. I tried by using Integration by parts but that didn't actually work out. Moreover, i would also be interested to see if there are other ways of solving this sum.