Another "Solvable" Series

[maverick280857]

Hi

This is another series

S = \sum_{i = 1}^{n} \frac{i^4}{(2i - 1)(2i + 1)}

I did it by dividing the numerator by the denominator (and doing the division once again) and finally using partial fractions. Is there a neater way? (I will post my solution to the problem in a few hours...going out right now).

Thanks a lot,

Cheers
Vivek

[Zurtex]

I've started by saying that:

\frac{i^4}{(2i-1)(2i+1)} = Ai^2 + Bi + \frac{C}{2i-1} + \frac{D}{2i + 1}

Once solved for A, B, C and D the sum becomes:

S = \left( A\sum_{i = 1}^{n} i^2 \right) + \left( B\sum_{i = 1}^{n} i \right) + \left( \sum_{i = 1}^{n} \frac{C}{2i-1} + \frac{D}{2i + 1} \right)

The former 2 sums are standard form, the latter one is a quite simple as it takes the form f(i) = g(i) - h(i)

That's the simplest way I see of doing it, but someone else might come up with a better method.

[maverick280857]

Hi Zurtex

That's EXACTLY what I did :biggrin:

Thanks anyway for clarifying my method. Yeah, lets see if someone comes up with something different.

Cheers
Vivek