evaluation of series Evaluate :
a.$$\frac{1}{2\cdot3}+\frac{1}{4\cdot5}+\frac{1}{6\cdot7}+\cdots$$
For this I looked at 
$$x+x^3+x^5+\cdots=\frac{x}{1-x^2} \text{ for }|x|<1$$
Integrating the above series from $0$ to $t$ yields
$$\frac{t^2}{2}+\frac{t^4}{4}+\frac{t^6}{6}+\cdots=\int_{0}^{t}\frac{x}{1-x^2} \, dx$$
Again integrating the series from $0$ to $1$ will give 
$$\frac{1}{2\cdot3}+\frac{1}{4\cdot5}+\frac{1}{6\cdot7}+\cdots=\int_{0}^{1}\left(\int_{0}^{t}\frac{x}{1-x^2} \, dx\right) \, dt$$
But upon integration $\log(1-t^2) $ comes which is not defined at $t=1$

Using $\displaystyle \frac1{n(n+1)}=\frac1n-\frac1{n+1}$
$$\frac1{2\cdot3}+\frac1{4\cdot5}+\frac1{6\cdot7}+\cdots=\frac12-\frac13+\frac14-\frac15+\frac16-\frac17+\cdots$$
Now use Convergence for log 2 or Taylor series for $\log(1+x)$ and its convergence
