Divergent Series question

[BrownianMan]

(1) Using the Archimedean definition of divergence, prove that if $\sum_{i=1}^{\infty }x_{i}$ diverges to infinity, then so does either $\sum_{i=1}^{\infty }x_{2i}$ or $\sum_{i=1}^{\infty }x_{2i+1}$.

(2) Show an example where $\sum_{i=1}^{\infty }x_{i}$ diverges to infinity but $\sum_{i=1}^{\infty }x_{2i}$ does not.

So this is what I have for (1):

If $\sum_{i=1}^{\infty }x_{i}$ diverges then for all N, there is a number m such that $N < \sum_{i=1}^{\infty }x_{i}$ diverges for all $k\geq m$. Then, $\lim_{k \to \infty }\left | \frac{x_{k+1}}{x_{k}} \right | > 1$ or $\lim_{k \to \infty }\left | \frac{x_{k+1}}{x_{k}} \right | = \infty$ and $\lim_{k \to \infty }\left | \frac{x_{k+1}}{x_{k}} \right |$ will become larger than 1 at some point. There exists an m such that $\left | \frac{x_{k+1}}{x_{k}} \right | > 1$ when $k\geq m$. This means that when $k\geq m$, $x_{k} < x_{k+1}$ and $\lim_{k \to \infty }x_{k} \neq 0$. $x_{k} < x_{k+1}$ implies that either $x_{k} < x_{2k}$ or $x_{k} < x_{2k+1}$, depending on whether or not the series is alternating. So either $\sum_{i=1}^{k }x_{i} < \sum_{i=1}^{k }x_{2i}$ or $\sum_{i=1}^{k }x_{i} < \sum_{i=1}^{k }x_{2i+1}$, and we have a number m such that $N < \sum_{i=1}^{k }x_{i} < \sum_{i=1}^{k }x_{2i}$ or $N < \sum_{i=1}^{k }x_{i} < \sum_{i=1}^{k }x_{2i+1}$ for all $k\geq m$. Therefore, $\sum_{i=1}^{\infty }x_{2i}$ or $\sum_{i=1}^{\infty }x_{2i+1}$ diverges.

Is this right?

I can't think of an example for (2). Any help?

 

[nonequilibrium]

I don't seem to understand the first line of your proof, but I think that's my fault: I've never heard of the "archimedean definition of divergence", so never mind me.

But I wanted to comment on (2): it's always handy to think of the most extreme case, i.e. one where $\sum x_{2i}$ is zero, what's an obvious candidate? Use the left-over freedom to construct a divergence sequence.

 

[BrownianMan]

Thanks.

Can anyone look over my proof and tell me if there are any logical errors?