Pi series

1. Dec 11, 2006

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

2. Dec 11, 2006

Hurkyl

Staff Emeritus
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.

3. Dec 11, 2006

the dude man

Sorry i think the (1/4) is positive

that should give (Pi/2)

yes/no?

4. Dec 11, 2006

the dude man

Sorry heres the correct series

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

5. Dec 12, 2006

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.

6. Dec 12, 2006

the dude man

That works but are there any other options?

Last edited: Dec 12, 2006
7. Dec 13, 2006

dextercioby

I don't see another one.

Daniel.

8. Dec 14, 2006

the dude man

Im interested in the ways you can obtain the series.