# Infinite series proof

1. Sep 29, 2011

### BrownianMan

Prove that $\sum_{i=1}^{\infty }x_{i}$ = $\sum_{i=1}^{\infty }(x_{2i} + x_{2i+1})$ if $\sum_{i=1}^{\infty }x_{i}$ converges and if for any $\varepsilon > 0$ there is some m such that $|x_{k}| < \varepsilon$ for all $k\geq m$.

I'm a little confused by this because for 1/(4^i) for i from 1 to infinity, this doesn't hold. The sum of 1/(4^2i) for i from 1 to infinity is 1/15 and the sum of 1/(4^(2i+1)) for i from 1 to infinity is 1/60. Together they have a sum of 1/12. But the sum of 1/(4^i) for i from 1 to infinity is 1/3. Doesn't this only hold when the index starts at i=0?

2. Sep 29, 2011

### SammyS

Staff Emeritus
There is a small problem with the indices. The sum on the right is missing x1.

3. Sep 29, 2011

### BrownianMan

Yes! That's exactly what I was thinking. So there must be something wrong with the question, right?

4. Sep 29, 2011

### SammyS

Staff Emeritus
It would work if the sums started at i = 0.

It would also work if the index 2i+1 were changed to 2i-1 in the sum on the right side, or if the sum on the left side had i start at 2, etc.

Of course the equality can only hold if both series converge.

5. Sep 29, 2011

### BrownianMan

True.

I'm not sure if I need to outline the different scenarios for which the equality holds, or just state that the equality as given in the problem is not true. I guess I'll have to ask my prof.

Thanks for the resposnes though.

6. Sep 29, 2011

### SammyS

Staff Emeritus
I suspect the both sums should start with i=0, & that's the version your prof. will want you work with.