The rule for the sum of this series?

[Saracen Rue]

TL;DR Summary

What is the rule for the following series: $$\frac {1}{1×3}+\frac {1}{5×7}+\frac {1}{9×11}...$$

Consider the following series with the following pattern $$\frac {1}{1×3}+\frac {1}{5×7}+\frac {1}{9×11}...$$

How would you go about working out what the general rule for this sum is? That is in the form of $\sum_{n=a}^{b}f(n)$

Any help is greatly appreciated.

 

Your sum is $$\sum_{n=0}^\infty \frac{1}{(4n+1)(4n+3)}$$

Do you have to evaluate it?

 

[Saracen Rue]

Your sum is $$\sum_{n=0}^\infty \frac{1}{(4n+1)(4n+3)}$$

Do you have to evaluate it?

Thank you so much, this had really been bugging me. I actually did try testing that mentally but I guess I messed something up.

 

[Svein]

Your sum is $$\sum_{n=0}^\infty \frac{1}{(4n+1)(4n+3)}$$

Do you have to evaluate it?

Continuing the processing: $\sum_{n=0}^{\infty}\frac{1}{(4n+1)(4n+3)}=\frac{1}{2}\sum_{n=0}^{\infty}(\frac{1}{4n+1}-\frac{1}{4n+3})=\frac{\pi}{8}$