summation - Find the sum of the series of $frac{1}{1cdot 3}+frac{1}{3cdot 5}+frac{1}{5cdot 7}+frac{1}{7cdot 9}+...$

Find the sum of the series

$$\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots$$
My attempt solution:
$$\frac13\cdot\left(1+\frac15\right)+\frac17\cdot\left(\frac15+\frac19\right)+\frac1{11}\cdot\left(\frac19+\frac1{13}\right)+\cdots$$
$$=\frac13\cdot\left(\frac65\right)+\frac17\cdot \left(\frac{14}{45}\right)+\frac1{11}\cdot\left(\frac{22}{117}\right)+\cdots$$
$$=2\cdot\left(\left(\frac15\right)+\left(\frac1{45}\right)+\left(\frac1{117}\right)+\cdots\right)$$

$$=2\cdot\left(\left(\frac15\right)+\left(\frac1{5\cdot9}\right)+\left(\frac1{9\cdot13}\right)+\cdots\right)$$
It is here that I am stuck. The answer should be $\frac12$ but I don't see how to get it. Any suggestions?

Also, a bit more generally, are there good books (preferably with solutions) to sharpen my series skills?

Answer

This is a general approach to evaluate the sum of series, like these.

First find $n^{th}$ term of series.

Let $T_n$ denote the $n^{th}$ term.

We see that,

$T_1 = \frac{1}{\color{green}{1} \cdot \color{teal}{3}}$

$T_2 = \frac{1}{\color{green}{3} \cdot \color{teal}{5}}$

And so on.
Let the numbers in $\color{green}{green}$ be

$$\color{green}{X_1,X_2,X_3,X_4,..=1,3,5,7...}$$

Clearly they form an A.P. with common difference $=2$

So, $n^{th}$ term of this AP is $1 + (n-1) × 2 = \color{green}{2n-1}$

Similarly,
Let the numbers in $\color{teal}{teal}$ be

$$\color{teal}{Y_1,Y_2,Y_3,Y_4,..=3,5,7,9...}$$
Clearly they form an A.P. with common difference $=2$

So, $n^{th}$ term of this AP is $3 + (n-1) × 2 =\color{teal}{ 2n+1 }$

So, the $n^{th}$ term of the main question is just

$$T_n = \frac{1}{\color{green}{(2n-1)} \cdot \color{teal}{(2n+1)}}$$

Now, taking summation from $1$ to $n$ , we have,

$$\sum_{n=1}^n \frac{1}{(2n-1) \cdot (2n+1)}$$

$$= \sum_{n=1}^n = \frac{1}{2} \cdot \frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}$$

$$= \sum_{n=1}^n = \frac{1}{2} \cdot \frac{1}{(2n-1)} - \frac{1}{(2n+1)}$$

$$= \sum_{n=1}^n = \frac{1}{2} \cdot ( 1 - \frac{1}{(2n+1)} )$$

While $n = ∞$,

$$\sum_{n=1}^∞ = \frac{1}{2} \cdot ( 1 - \frac{1}{(2(∞)+1)} )$$

$$= \sum_{n=1}^∞ = \frac{1}{2} \cdot ( 1 - \frac{1}{∞} )$$

Since $\frac{1}{∞} = 0$,

$$\sum_{n=1}^∞ = \frac{1}{2} \cdot ( 1 - 0 )$$

Which is

$$\sum_{n=1}^∞ = \frac{1}{2}$$