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I have to investigate convergence of series $$\sum_{k=10}^{+\infty}\frac{(-1)^k}{k+(-1)^k}$$ It certainly does not converge absolutely, because it is basically a harmonic series with every two elements flipped and if harmonic series converged, or greater than series $$\sum_{k=10}^{+\infty}\frac{1}{k-1}$$ However, I do not know, how to prove convergence of this series overall. I thought about the fact, that it is basically a harmonic series(well, alternating harmonic or whatever I should call it) with every two elements flipped and if normal "alternating harmonic series" converges, then this rearangement could technically converge too. However, I am not sure if this is sufficient enough and if I can do this and I am afraid that it is not. I was considering those theorems that say "every divergent series with alternating elements can be rearranged to make arbitrary finite sum" and I am not sure if it holds with convergent series, respectively, if it holds against convergent series and infinity - that you can rearrange convergent series to make divergent...

Neverthless, there should be an easy way to prove that this sum is convergent/divergent, since it is in our textbook.

Also I thought, since it is certainly lesser or equal to $1/(k-1)$ and greater or equal to $1/(k+1)$ assigning numbers $S_1$ and $S_2$ to sums of these "alternating harmonics" and just saying that it is lower bounded by $S_2$ and upper bounded by $S_1$, since those series are convergent and have to be upper and lower-bounded too. But also, I am not sure if this is enough.

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  • $\begingroup$ Look at the sum of two consecutive terms. $\endgroup$ – Daniel Fischer♦ Jun 22 '14 at 1:14
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Hint: Note that $$ \sum_k\frac{(-1)^k}{k+(-1)^k}=\sum_k\frac{(-1)^k}{k}-\sum_k\frac{1}{k\cdot(k+(-1)^k)} $$ and prove that the last series on the right has positive terms and converges (absolutely).

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By grouping consecutive terms (we can do this since the terms tend to 0), the series can be rewritten as:

$\sum_{k=5}^\infty(\frac{1}{2k + 1} - \frac{1}{2k}) = \sum_{k=5}^\infty\frac{-1}{2k(2k + 1)} > \sum_{k=5}^\infty\frac{-1}{(2k)^2}$

So the sequence converges by comparison.

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    $\begingroup$ (for the sake of getting the fine details right, this argument also uses the fact the terms converge to zero) $\endgroup$ – user14972 Jun 22 '14 at 2:04
  • $\begingroup$ How do you mean? $\endgroup$ – Mathmo123 Jun 22 '14 at 2:05
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    $\begingroup$ To see the problem, note that grouping consecutive terms would imply $\sum_{i=0}^{\infty} (-1)^n = 0$. You need to have terms decreasing to zero to show that all of the partial sums of the original series are near the partial sums of the new sum. $\endgroup$ – user14972 Jun 22 '14 at 2:06
  • $\begingroup$ Ah I see. Thanks. $\endgroup$ – Mathmo123 Jun 22 '14 at 2:07

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