3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @ParclyTaxel Gah! My inability to read signs when strikes again. Retracted. $\endgroup$ – Xander Henderson Dec 8 '17 at 3:22
  • $\begingroup$ Your question suggests an Infinite sum. Am I right? $\endgroup$ – Ravi Prakash Dec 8 '17 at 5:26
4
$\begingroup$

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

$$ \sum_{n=1}^∞ = \frac{1}{2} $$

$\endgroup$
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

\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 $n\rightarrow\infty$.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.