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

I can see that if you group the terms in pairs of two, you get $(1-\frac{1}{2})+(\frac{2}{3}-\frac{1}{3})+(\frac{2}{4}-\frac{1}{4})+(\frac{2}{5}-\frac{1}{5})$...=$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}$...

The partial sums $S_{2n}$= the harmonic series from n=2 to n.

So I intuitively would guess the series diverges. I have done a lot of problems in my real analysis class that involved this kind of thinking, but can not find anything about it or what it is called and I would like to know more.

I am very concerned that the "speed" at which I compare one series to another will have an affect on the convergence/divergence.

  • $\begingroup$ You can use the associativity only if the series converges absolutely. In your example the series does not converge absolutely. $\endgroup$ – Monolite Apr 15 '18 at 21:10
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    $\begingroup$ @Monolite: conditional convergence is not a problem for associativity because you are still adding the terms in the same order. Commutativity is a problem for conditional convergence because you can add the positive terms much more quickly than the negative terms. $\endgroup$ – Ross Millikan Apr 15 '18 at 21:21
  • $\begingroup$ @RossMillikan Thanks! $\endgroup$ – Monolite Apr 15 '18 at 21:27
  • $\begingroup$ $1/2+1/3+1/4+1/5+1/6+1/7+1/8+......\geq$ $1/2 +(1/4+1/4)+(1/8+1/8+1/8+1/8)+...=$ $1/2+1/2+1/2+...=\infty.$ $\endgroup$ – DanielWainfleet Apr 15 '18 at 21:58
  • $\begingroup$ Have you seen "telescoping" (finite or infinite) series? $\endgroup$ – DanielWainfleet Apr 15 '18 at 22:03

A series $\sum_n a_n$ converges, by definition, if and only if the sequence of it's partial sums $(A_n)_n$ is convergent.

Here, you have that $A_{2n} = H_n - 1 \xrightarrow[n\to\infty]{}\infty$. Therefore, the sequence $(A_n)_n$ is not convergent (as otherwise every of its subsequences would be).


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