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Does the following variant of the harmonic series converge? If it diverges (which I think it does), can I know if it diverges to $\infty$ or has no limit?

Note that the series is not alternating in the classical sense of the word. $$1 + \frac{1}{2} -\frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots$$

The generic term of the series would have to be something like,

$$a_n = \left\{\begin{array}{ll} -\frac{1}{n}, & \text{if } 3 \mid n \\ \;\, \,\, \frac{1}{n}, & \text{otherwise}\end{array}\right.$$ I'm not sure if it's very helpful. The terms divisible by 3 are negative, and the others are positive.

Is there a way to decide and prove whether an alternating series of this sort (e.g. with a period other than 2) converges? Or one where terms are positive or negative according to some other rule?

Almost all convergence tests I've come across are generally limited to either simple alternating series or where all the terms are positive.

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  • $\begingroup$ what is the generic term in this series? $\endgroup$ – Jonathan Aug 10 '13 at 1:30
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    $\begingroup$ Nope. Are you familiar with asymptotics of harmonic numbers? $\endgroup$ – anon Aug 10 '13 at 1:30
  • $\begingroup$ This series actually converges by the alternating series test. This is a theorem that states that if an infinite series is alternating, then it converges iff the $n^{\text{th}}$ term goes to $0$ as $n$ goes to $\infty$. $\endgroup$ – ithisa Aug 10 '13 at 1:34
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    $\begingroup$ It's bigger than $1+(1/4)+(1/7)+(1/10)+\dots$ which, of course, diverges. $\endgroup$ – Gerry Myerson Aug 10 '13 at 1:34
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    $\begingroup$ I think the general rule for $\sum a_n/n$, where the $a_n$ are $\pm1$, is that if the signs are periodic then the sum converges if and only if there are the same number of plus and minus in each period. $\endgroup$ – Gerry Myerson Aug 10 '13 at 1:49
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No, the series diverges. Note that the partial sum of length $3N$ can be written as

$$\sum_{n=0}^{N-1} \left(\frac{1}{3n+1} + \frac{1}{3n+2} - \frac{1}{3(n+1)}\right).$$

Combining the terms, we get

$$\sum_{n=0}^{N-1} \frac{9 n^2+18 n+7}{3 (n+1) (3n+1) (3 n+2)}$$

and it's easy to see that this is a divergent series by limit comparison with $\sum 1/n$. Now, if the original series converged, then this sub-sequence of the sequence of partial sums would also have to converge.

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    $\begingroup$ Thank you! That's a great answer and it helped me understand the general case as well. $\endgroup$ – GregRos Aug 10 '13 at 15:48
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Note that the partial sums $S_n \geq 1 + \frac{1}{4} + \frac{1}{7} + \ldots = \sum_{k=0}^{[n/3]}\frac{1}{3k+1}$, bounded from below by some partial sum of the harmonic series (divided by $3$), which diverge to infinity.

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  • $\begingroup$ Gerry Myerson, you had me. Apologies for mis-quoting your name. $\endgroup$ – Jonathan Y. Aug 10 '13 at 1:51
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It's quite easy to see it diverges. It is the difference between two series, 1/n and 2/(n(n+1)). The first diverges, the second converges.

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  • $\begingroup$ the pattern isn't triangular, it's just divisible by $3$ $\endgroup$ – MCT Feb 27 '15 at 17:37

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