High School The rule for the sum of this series?

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The series discussed is of the form $$\sum_{n=0}^\infty \frac{1}{(4n+1)(4n+3)}$$. Participants explore how to derive a general rule for this sum. The evaluation of the series leads to the conclusion that it equals $$\frac{\pi}{8}$$. The discussion emphasizes the importance of correctly applying series manipulation techniques. Overall, the series converges to a specific value, providing a solution to the initial query.
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TL;DR
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.
 
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Your sum is $$\sum_{n=0}^\infty \frac{1}{(4n+1)(4n+3)}$$

Do you have to evaluate it?
 
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Math_QED said:
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.
 
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Math_QED said:
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}
 
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