# Thread: A problem involving sums.

1. ## A problem involving sums.

The following question refers to http://www.mathhelpforum.com/math-he...-162658-3.html- a post of Bruno J. to chiph588@ problem.

I will be glad if someone could explain me why $\sum_{n=0}^\infty \frac1{\alpha^{n-m}+1}=\sum_{n=1}^m \frac1{\alpha^{-n}+1}+F(1)=\sum_{n=1}^m \frac1{\alpha^{-n}+1}+\sum_{n=0}^\infty \frac{1}{\alpha^n+1}$

Thank you!!!

2. Just notice $\sum _{n=0}^{\infty} \frac{1}{\alpha ^{n-m}+1}=\sum _{n=0}^{m} \frac{1}{\alpha ^{n-m}+1}+\sum _{n=m+1}^{\infty} \frac{1}{\alpha ^{n-m}+1}$ and now just make k=n-m which gives you the equality between the first and last, and the rest follows from the definition of F

3. Originally Posted by Jose27
Just notice $\sum _{n=0}^{\infty} \frac{1}{\alpha ^{n-m}+1}=\sum _{n=0}^{m} \frac{1}{\alpha ^{n-m}+1}+\sum _{n=m+1}^{\infty} \frac{1}{\alpha ^{n-m}+1}$ and now just make k=n-m which gives you the equality between the first and last, and the rest follows from the definition of F
Alright... but... why the second sum starts from 0(to infinity)? (It should be 1, no?)

4. Originally Posted by Also sprach Zarathustra
Alright... but... why the second sum starts from 0(to infinity)? (It should be 1, no?)
I suggest you read very carefully what I wrote and what you're asking. Isn't the first term off too?

5. Originally Posted by Jose27
I suggest you read very carefully what I wrote and what you're asking. Isn't the first term off too?