- #1
BrownianMan
- 134
- 0
Prove that [itex]\sum_{i=1}^{\infty }x_{i}[/itex] = [itex]\sum_{i=1}^{\infty }(x_{2i} + x_{2i+1})[/itex] if [itex]\sum_{i=1}^{\infty }x_{i}[/itex] converges and if for any [itex]\varepsilon > 0[/itex] there is some m such that [itex]|x_{k}| < \varepsilon[/itex] for all [itex]k\geq m[/itex].
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?
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?