$\sum a_n$ converges conditionally

Question

If we assume that $\sum a_n$ converges conditionally then How can we comment that $\sum a_{2n} $ does not converges, While it does when $\sum a_n$ converges absolutely ?

Answer 1

For a counter-example (provided the sequence conditionally converges)... think of your favorite conditionally convergent sequence, the alternating harmonic series.

If you'd rather not have to show that the series containing only the positive terms of the alternating harmonic series diverges, you could use something more fun like the following:

$$ \frac{1}{1} - \frac{1}{1} + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + \frac{1}{4} - \frac{1}{4} + \frac{1}{5}-\frac{1}{5}+...$$

If you only take every other term, you have the harmonic series (diverges). But if you take all the terms, then the series converges to zero (values in the sequence of partial sums are either $\frac{1}{n}$ or 0 depending on how many terms you've summed).

Answer 2

It may or may not converge. Two examples are $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$ in which $\Sigma a_{2n}$ diverges, or $1 + 0 - \frac{1}{2} + 0 + \frac{1}{3} + 0 - \frac{1}{4}$, in which $\Sigma a_{2n}$ converges.

In the absolute convergence case, it's straightforward to see that $\Sigma a_{2n}$ converges absolutely if $\Sigma a_n$ does. One method of proof is the observation that a sequence converges absolutely if and only if the partial sums $\Sigma_{n=1}^k |a_n|$ are bounded above. Any upper bound for $\Sigma_{n=1}^k |a_n|$ is then an upper bound for $\Sigma_{n=1}^k |a_{2n}|$. This similarly shows that you can take any subsequence, not just the even terms.

An intuitive justification for the conditionally convergent case is that conditionally convergent series can have their convergence behavior altered by rearranging terms. Discarding the odd terms entirely is considerably more drastic, so there is no reason to believe that convergence will be preserved.