Exploring an Infinite Series: Pi/2?

[the dude man]

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

 

[Hurkyl]

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

$$\sum_{n = 1}^{+\infty} \frac{1}{ (n-1)^2 - \frac{1}{4}}$$

right? Anyways, the usual trick to these is to use partial fractions.

 

[the dude man]

Sorry i think the (1/4) is positive

that should give (Pi/2)

yes/no?

 

[the dude man]

Sorry here's the correct series

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

 

[dextercioby]

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.

 

[the dude man]

That works but are there any other options?

 

[dextercioby]

I don't see another one.

Daniel.

 

[the dude man]

Im interested in the ways you can obtain the series.