Finding the sum $\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\cdots$

Question

The sum of series is

$${1\over(3\times5)}+{1\over(5\times7)}+{1\over(7\times9)}+\cdots$$

I used to solve this problem as its $n$th term $${1\over (2n+1)(2n+3)}$$

Now how can I proceed??

Answer 1

This is a telescoping series: $$ \frac{1}{(2n+1)(2n+3)}=\frac{1}{2}\frac{(2n+3)-(2n+1)}{(2n+1)(2n+3)}=\frac{1}{2}\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right) $$ So: $$ \sum_{n=1}^N\frac{1}{(2n+1)(2n+3)}=\frac{1}{2}\sum_{n=0}^N\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)=\frac{1}{2}\left(\frac{1}{2\cdot 1+1}-\frac{1}{2N+3}\right) $$ In the limit, we get: $$ \sum_{n=1}^\infty\frac{1}{(2n+1)(2n+3)}=\frac{1}{6} $$

Answer 2

$$\frac{1}{(2k+1)(2k+3)}=\frac12\left(\frac1{2k+1}-\frac1{2k+3}\right)$$ $$\begin{align} \sum_{k=1}^n\frac{1}{(2k+1)(2k+3)} &=\frac12\sum_{k=1}^n\left(\frac1{2k+1}-\frac1{2k+3}\right)\\ &=\frac12\left(\left(\frac13-\frac15\right)+\left(\frac15-\frac17\right)+\dots+\left(\frac1{2n+1}-\frac1{2n+3}\right)\right)\\ &=\frac12\left(\frac13-\frac1{2n+3}\right)\\ &=\frac16-\frac1{2(2n+3)}\\ \end{align}$$ $$\therefore\sum_{k=1}^\infty\frac{1}{(2k+1)(2k+3)}=\lim_{n\to\infty}\left(\frac16-\frac1{2(2n+3)}\right)=\frac16$$

Answer 3

hint: Use that $$\frac{1}{(2n+1)(2n+3)}=\frac{1}{2(1+2n)}-\frac{1}{2(3+2n)}$$

Answer 4

Hint : The $n$ th term of your series is $$ a_n= \dfrac{1}{2}\left ( \dfrac{1}{2n+1}- \dfrac{1}{2n+3}\right)$$ expanding your series you get $$ \dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}...\right)$$ as you see the terms cancel out each other the only term left is $\dfrac{1}{3}$ with a factor of $\dfrac{1}{2}$ outside, giving your sum of the series as $\dfrac{1}{6}$

Answer 5

Let there are functions $$f(x) = \sum_{n=1}^\infty \frac{x^n}{\alpha n + \beta}$$

and $$\psi(x) = \frac{f(x,\alpha,\beta) - f(x,\alpha,\alpha + \beta)}{\alpha} = \sum_{n=1}^\infty \frac{x^n}{(\alpha n + \beta)(\alpha (n+1) + \beta)}$$

where $\alpha,\beta > 0$ and $x \in [-1,1]$.

Thereby $$\psi(x) = \frac{1}{\alpha}\left(\frac{x}{\alpha + \beta} + \lim_{n\to \infty} \frac{x^n}{\alpha(n+1) + \beta}\right) = \frac{x}{\alpha(\alpha + \beta)} .$$

For instance that $\alpha = 2,\beta = 1$ and $x = 1$, $$\psi(1) = \frac{1}{6}.$$