# 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?

• @ParclyTaxel Gah! My inability to read signs when strikes again. Retracted. – Xander Henderson Dec 8 '17 at 3:22
• Your question suggests an Infinite sum. Am I right? – Ravi Prakash Dec 8 '17 at 5:26

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

$$\sum_{n=1}^∞ = \frac{1}{2}$$

This is a classic telescoping series. $$\frac1{n\cdot(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right)$$ Thus $$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\frac{1}{9\cdot 11}+\cdots$$ $$=\frac12\left(\frac11-\frac13+\frac13-\frac15+\frac15-\frac17+\frac17-\frac19+\frac19-\frac1{11}+\cdots\right)$$ $$=\frac12$$

\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 $n\rightarrow\infty$.

$$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\frac{1}{9\cdot 11}...=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}\right)+\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\cdots=\dfrac12$$

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 $\psi(\frac{1}{2})-\psi(\frac{3}{2})=-2$