# Showing two infinite sums are equal?

• March 15th 2009, 07:00 AM
KZA459
Showing two infinite sums are equal?
I have this problem requiring me to show two infinite sums are equal but I can't seem to figure out how to do this. No matter what I try never give any results.

Hence my question is what approach should I use toward one of these question?

This is the question in question if it can help

Show
$
\sum_{p=0}^{\infty}\frac{x^{2p}}{1+x^{4p+2}}=\sum_ {q=0}^{\infty} (-1)^{q}\frac{x^{2q}}{1-x^{4q+2}}
$

if |x|<1
• March 15th 2009, 11:16 AM
Opalg
Quote:

Originally Posted by KZA459
I have this problem requiring me to show two infinite sums are equal but I can't seem to figure out how to do this. No matter what I try never give any results.

Hence my question is what approach should I use toward one of these question?

This is the question in question if it can help

Show
$
\sum_{p=0}^{\infty}\frac{x^{2p}}{1+x^{4p+2}}=\sum_ {q=0}^{\infty} (-1)^{q}\frac{x^{2q}}{1-x^{4q+2}}
$

if |x|<1

Use the binomial series for $(1+x^{4p+2})^{-1}$ to see that $\sum_{p=0}^{\infty}\frac{x^{2p}}{1+x^{4p+2}} = \sum_{p=0}^{\infty}x^{2p}\sum_{q=0}^{\infty}(-1)^qx^{(4p+2)q} = \sum_{p=0}^{\infty}\sum_{q=0}^{\infty}(-1)^qx^{4pq+2p+2q}$. Reverse the order of summation (why is this justified?) and then unwind the double sum to get $\sum_{q=0}^{\infty} (-1)^{q}\frac{x^{2q}}{1-x^{4q+2}}$.
• March 15th 2009, 12:42 PM
KZA459
Darn, thanks a lot Opalg.
• March 15th 2009, 05:05 PM
KZA459
Sorry, me again.

I also have the show a similar result for |x|>1

However using the power serie of $\frac{1}{1+x^{4p+2}}$
does not work in this case since it has as radius of convergence 1.

I must show if |x|>1 then
$
\sum_{p=0}^{\infty}\frac{x^{2p}}{1+x^{4p+2}}= -\sum_{q=0}^{\infty} (-1)^{q}\frac{x^{2q}}{1-x^{4q+2}}
$

Unfortunately I am still as stuck and can't find a way to figure that one out even with the insight you gave me Olpag.

Anyone has an idea?
• March 16th 2009, 03:40 AM
Opalg
Quote:

Originally Posted by KZA459
I also have the show a similar result for |x|>1

However using the power serie of $\frac{1}{1+x^{4p+2}}$
does not work in this case since it has as radius of convergence 1.

I must show if |x|>1 then
$
\sum_{p=0}^{\infty}\frac{x^{2p}}{1+x^{4p+2}}= -\sum_{q=0}^{\infty} (-1)^{q}\frac{x^{2q}}{1-x^{4q+2}}
$

Unfortunately I am still as stuck and can't find a way to figure that one out

If you want a series that converges when |x| > 1 then you need it to use negative powers of x. So write $\frac{x^{2p}}{1+x^{4p+2}}= x^{-2}\frac{(x^{-1})^{2p}}{1+(x^{-1})^{4p+2}}$ (dividing top and bottom by $x^{4p+2}$). You can then apply the previous result with $x^{-1}$ in place of x, noticing that an extra minus sign will come in when you transform it back into positive powers of x.
• March 16th 2009, 11:13 AM
KZA459
Thanks :)