Is the Infinite Series of Pi/2 a Mathematical Marvel?

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SUMMARY

The forum discussion centers on the infinite series representation of Pi/2, specifically the series defined as Σ (1 / ((2n-1)² - 1/4)) from n=1 to infinity. Participants confirm that using partial fractions is a valid method for deriving this series, leading to the conclusion that it sums to Pi/2. The correct formulation of the series is emphasized, and the importance of accurate notation is highlighted. The discussion also invites exploration of alternative methods for obtaining the series.

PREREQUISITES
  • Understanding of infinite series and convergence
  • Familiarity with partial fraction decomposition
  • Basic knowledge of mathematical notation and summation
  • Experience with mathematical software like Maple
NEXT STEPS
  • Research the proof of the series Σ (1 / ((2n-1)² - 1/4)) converging to Pi/2
  • Learn about alternative methods for deriving infinite series
  • Explore the application of partial fractions in calculus
  • Investigate the use of Maple for mathematical proofs and series analysis
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Mathematicians, educators, and students interested in series convergence, mathematical proofs, and the applications of partial fractions in calculus.

the dude man
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Hey guys!

My friend showed me a infinite series on maple

infinity
---
\ ( ((n-1)^2) - (1/4) )^(-1) = (Pi/2)
/
---
n=1

Is anyone familar with this?
If so can you refer me to its proof? Or maybe post it.

Thanks
 
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Hrm. Unless I made a mistake, that sums to 2. You meant

<br /> \sum_{n = 1}^{+\infty} \frac{1}{ (n-1)^2 - \frac{1}{4}}<br />

right? Anyways, the usual trick to these is to use partial fractions.
 
Sorry i think the (1/4) is positive

that should give (Pi/2)

yes/no?
 
Sorry here's the correct series

infinity
----
\ (((2*n-1)^2) - (1/4))^(-1)
/
----
n=1
 
Use partial fractions, as hinted already.

\frac{1}{\left(2n-1\right)^{2}-\frac{1}{4}}=2\left(\frac{1}{4n-3}-\frac{1}{4n-1}\right)

Daniel.
 
That works but are there any other options?
 
Last edited:
Im interested in the ways you can obtain the series.
 

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