Find the sum of the series of $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+...$ 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?
 A: 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
$$ \sum_{n=1}^∞ = \frac{1}{2} $$
A: This is a classic telescoping series.
$$\frac1{n\cdot(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right)$$
Thus
$$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\frac{1}{9\cdot 11}+\cdots$$
$$=\frac12\left(\frac11-\frac13+\frac13-\frac15+\frac15-\frac17+\frac17-\frac19+\frac19-\frac1{11}+\cdots\right)$$
$$=\frac12$$
A: \begin{align*}
\sum_{n=1}\dfrac{1}{(2n-1)(2n+1)}&=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\
&=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{2n-1}-\dfrac{1}{2(n+1)-1}\right)\\
&=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{f(n)}-\dfrac{1}{f(n+1)}\right)\\
&=\dfrac{1}{2}\dfrac{1}{f(1)},
\end{align*}
where $f(n)=2n-1$, and note that $f(n)^{-1}\rightarrow 0$ as $n\rightarrow\infty$.
A: $$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\frac{1}{9\cdot 11}...=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}\right)+\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\cdots=\dfrac12$$
A: Let me give a general method which is useful 
for this sum $\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots=\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}$
we have 
$\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}=\sum_{n=0}^\infty\dfrac{1}{(2n+1)(2n+3)}$
we can use this method 
$$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{1}$$
where $\psi$ is digamma function.
now we can write 
$\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}= \frac{1}{4}\sum_{n=1}^\infty\dfrac{1}{(n+\frac{1}{2})(n+\frac{3}{4})}=\frac{1}{2}$
since $\psi(\frac{1}{2})-\psi(\frac{3}{2})=-2$
