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.
Hint

By $(i)$ we have
$$\frac{x}{(1+(n-1)x)(1+nx)}=\frac{1}{1+(n-1)x}-\frac{1}{1+nx}$$
and note that if $x&gt;0$, then$$\sum_{n=1}^N\left(\frac{1}{1+(n-1)x}-\frac{1}{1+nx}\right)$$ is a telescoping sum.  Then, you need to show that $$\lim_{N\to\infty}\frac{N}{1+Nx}&lt;\infty$$ for all $x\ne 0$ and find the limit.
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, $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}+\frac{x}{(1+2x)(1+3x)}+\frac{x}{(1+3x)(1+4x)}+.....+\frac{x}{(1+(N-1)x)(1+Nx)}=\frac{N}{1+Nx}$