Does Splitting a Divergent Series Affect Its Infinity Sum?

[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?