Summation of the telescoping series $\sum_{n=1}^N \frac{x}{(1+(n-1)x)(1+nx)}$ $(i)$ Verify that $$\frac{1}{1+(n-1)x} - \frac{1}{1+nx} = \frac{x}{(1+(n-1)x)(1+nx)}$$
$(ii)$ Hence show that for $x  \ne 0$, $$\sum_{n=1}^N \frac{x}{(1+(n-1)x)(1+nx)}=\frac{N}{1+Nx}$$
Deduce that the infinite series $\frac{1}{1.\frac{3}{2}} +\frac{1}{2.\frac{3}{2}} +\frac{1}{2.\frac{5}{2}}.....$ is convergent and state the sum to infinity.
I need some help with the last part. I tried using the identity in the first part by letting $x=1$ and then varying $n$ but that did not produce the series that they have given. I also tried the same thing after setting $x=2$ but again could not get the series that they have given. 

From the LHS (The series), you can see that for:
At n=1, we have $\frac{x}{1+x}$
At n=2, we have $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}=\frac{2}{1+2x}$ (By simplifying)
At n=3, we have $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}+\frac{x}{(1+2x)(1+3x)}=\frac{3}{1+3x}$ (Again by simplifying)
At n=4, we have  $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}+\frac{x}{(1+2x)(1+3x)}+\frac{x}{(1+3x)(1+4x)}=\frac{4}{1+4x}$ (You see the partern?)
Therefore, By mathematical induction, we have......
At n=N, $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}+\frac{x}{(1+2x)(1+3x)}+\frac{x}{(1+3x)(1+4x)}+.....+\frac{x}{(1+(N-1)x)(1+Nx)}=\frac{N}{1+Nx}$
