Say I have: $(x_n)$ a sequence of real numbers such that $\sum x_n$ which converges conditionally and implies $\sum x_{2n}$ diverges.

I want to show that $x_{2n}$ does not in general converge.

So I have picked out a candidate for $\sum x_n$ = $\sum \frac{-1^{n+1}}{n}$

and so $\sum x_{2n} = \sum \frac{-1^{2n+1}}{2n}$

How can I argue rigorously that the first series converges and the second one diverges? I am aware that these both share the properties.

For the first series, I used an alternating series test,

since if we let $(b_n) = \frac{1}/{n}$ Then we see that $b_1 \geq b_2 \geq b_3 \dots $

and also that $(b_n) \to 0$ So then $ \sum (-1)^{n+1} b_n $ converges.

But how do I show divergence for the second series?


1 Answer 1


It's one half of the (negative of the) harmonic series. Whatever you preferred method of showing the harmonic series diverges (often the integral test or by directly showing the limit is infinite) works here.

  • $\begingroup$ We wrote the same thing, but you did it first :) sorry, I delete my answer. $\endgroup$ Apr 15, 2015 at 19:35

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