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.

  • $\begingroup$ Look at the sum of two consecutive terms. $\endgroup$ – Daniel Fischer Jun 22 '14 at 1:14
  • $\begingroup$ The infinite sum starting at $k=2$ is given by $\log(2) - 1 \simeq -0.306853$ $\endgroup$ – Dr. Wolfgang Hintze Jun 4 '20 at 7:49

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).


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.

  • 1
    $\begingroup$ (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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