Compute the sum

[cateater2000]

Compute the following sum
when n goes from 1 to oo of
1/(n^2(2n-1))


So far this is what i've done
1/(n^2(2n-1))=4/(2n(2n-1))-1/n^2
I know the sum from 1 to oo of 1/n^2 =(pi^2)/6
So I get
1/(n^2(2n-1))=4/(2n(2n-1)) - pi^2/6

so I need to compute the sum of 4/(2n(2n-1)) which i have no idea how to do.

Any hints or tips would be great

[TenaliRaman]

Assuming u did the first part correctly,
for ur question,
use partial fractions,
4/((2n)(2n-1))
= (4/(2n-1))-(4/(2n))

Now write the first few numbers of the series in this fashion and see if can notice something peculiar ...

-- AI

[cateater2000]

thankyou the reason I couldn't do the question is because I went
4/(2n(2n-1))=4/(2n-1)-2/n

and I couldn't see any apparent pattern.

I will try your partial fraction right now. Thankyou

[stunner5000pt]

by teh way i was wondering if i got your question right is it 1 / n ^(4n-2)

or is it 1/n^2 * (2n-1)

in any case the first one does converge since n^2 / n^4n < n^2 / n^4 = 1 / n^2 which is a convergent p series
the second one you can use the limit comparison test to show that it is lesser the 1/n^2 which is a convergent p series