The rule for the sum of this series?

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Discussion Overview

The discussion revolves around the evaluation of a specific infinite series of the form $$\frac {1}{1×3}+\frac {1}{5×7}+\frac {1}{9×11}...$$ Participants are exploring how to express this series in a general summation form and whether it can be evaluated to a specific value.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant presents the series and seeks a general rule for its sum in the form of $$\sum_{n=a}^{b}f(n)$$.
  • Another participant identifies the series as $$\sum_{n=0}^\infty \frac{1}{(4n+1)(4n+3)}$$ and questions whether it needs to be evaluated.
  • A subsequent reply reiterates the identification of the series and expresses gratitude for the clarification, indicating some confusion in their earlier mental calculations.
  • Another participant continues the evaluation process, suggesting that the series can be expressed as $$\frac{1}{2}\sum_{n=0}^{\infty}(\frac{1}{4n+1}-\frac{1}{4n+3})$$ and claims it equals $$\frac{\pi}{8}$$.

Areas of Agreement / Disagreement

Participants generally agree on the form of the series, but there is no consensus on the necessity or correctness of the evaluation process, as well as the final value proposed.

Contextual Notes

There may be limitations in the assumptions made regarding convergence and the evaluation steps, which are not fully resolved in the discussion.

Saracen Rue
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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: [itex]\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}[/itex]
 
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