Handling different summation limits

My question is to understand the difference of summation limits based on these example distributions.

Let's say that $x = 0, 1, 2, ..., 2m$

Then we have $$\frac{\delta^x}{1+\delta^{2m+1}}$$ When taking the summation of this distribution I would do: $$\sum_{x=0}^{2m} \frac{\delta^x}{1+\delta^{2m+1}}$$

However, let's change the variable and replace it with z when $0 < z < 2m$, I get this distribution: $$\frac{\delta^z}{1+\delta^{2m+1}}$$

But what happens to the limits of the summation? Surely, $0 < z < 2m$ is different to when $x = 0, 1, 2, ..., 2m$

The best I can get is: $$\sum_{0<z<2m}\frac{\delta^z}{1+\delta^{2m+1}}$$

However, I wouldn't know how to compute this sum or get the right lower and upper limits.

Solution 1:

You are overthinking this question, and relying too much on the formal summation expression. Write the sums out using ellipses and look at the patterns.

In each of these two sums the denominator of each term is the same, so you can just divide by it when you sum the numerators that vary.

To evaluate the first sum you need $$ 1 + \delta + \delta^2 + \cdots + \delta^{2m-1} + \delta^{2m} . $$

You can find it using the standard formula for the sum of a geometric series.

For the second sum the numerators add this way: $$ \delta + \delta^2 + \cdots + \delta^{2m-1} . $$

To calculate that, just factor out $\delta$ and sum the resulting ordinary geometric series that starts with exponent $0$ and ends with exponent $2m-2$.