# Spliting into even and odd $n$ sum

In this site, on proof 1 section there is a line where the expression is split into even and odd $n$, I don't understand how this was done, could someone explain it for me?

• Write out the entire sum, then pretend each pair of terms is one term of another sum. – Element118 Nov 29 '15 at 1:40

## 1 Answer

What it did was that it changed this original sequence $$\sum_{m=0}^{\infty}a_m=a_0+a_1+a_2+...$$ into $$\sum_{n=0}^{\infty}(a_{2n}+a_{2n+1})=(a_0+a_1)+(a_2+a_3)+...$$ by showing two adjacent terms at the same time.

So $n=0$ in the new sequence would correspond to $m=0$ & $1$ in the old sequence, and $n=1$ corresponds to $m=2$ & $3$, and $n=2$ to $m=4$ & $5$, and so on.

This way, $(-1)^m$ can be split into two cases - $m$ is even and $m$ is odd, and can be explicitly calculated.

• Thank you. I didn't think it was so simple – João Nov 29 '15 at 13:59