I do not understand why $$\sum^{\infty}_{k=1} z_k = \sum^{\infty}_{k=1} \frac1k$$ is divergent but the other series $$\sum^{\infty}_{k=1} z_k = \sum^{\infty}_{k=1} \frac{(-1)^{k+1}}k$$ is convergent. For both cases the $\displaystyle\lim_{n \to +\infty} z_{n} = 0$. Could you explain please?

I prefer to come up with an all-inclusive test of convergence that accepts the second but rejects the first. Thanks.

  • 1
    $\begingroup$ It diverges because of this. The fact that $\lim z_n = 0$ is only useful in that we can't immediately conclude divergence in either case. $\endgroup$
    – user61527
    Jul 28, 2014 at 22:04
  • $\begingroup$ Yes, to keep generality, let's assume we're in complex plane. $\endgroup$
    – Elnaz
    Jul 28, 2014 at 22:06
  • $\begingroup$ Complex plane or real, it doesn't matter, the first diverges because the little bits after a long while add up to a lot, while cancellation keeps the partial sums small in the second series. $\endgroup$ Jul 28, 2014 at 22:11
  • $\begingroup$ for the second series use the Leibnz criterion $\endgroup$ Jul 28, 2014 at 22:11

4 Answers 4


If the series $\sum_{n=1}^{\infty} \frac{1}{n}$ would converge,from the Cauchy criterion,there would be a $n_0 \in \mathbb{N}$ such that $\forall m>n \geq n_0:$ $$\frac{1}{n+1}+\frac{1}{n+2}+ \dots +\frac{1}{m}<\frac{1}{2}$$




$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n+n} \geq \frac{1}{n+n} +\frac{1}{n+n} \dots+\frac{1}{n+n} =\frac{n}{2n}=\frac{1}{2}$$

Therefore,the series diverges.


To conclude that the series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converges,we use the Dirichlet criterion:

Let $a_n,b_n$ two sequences of real numbers such that

  1. The sequence of the partial sums of the serie $\sum_{n=1}^{\infty} a_n$ is bounded

  2. $b_n$ is decreasing and converges to $0$

Then,the series $\sum_{n=1}^{\infty} a_nb_n$ converges.

In your case,take:

$$b_n=\frac{1}{n} \text{ and } a_n=(-1)^{n+1}$$


The approach I would use would be to expand the sequences and see whether or not a lower or upper limit can be placed.

For $\frac1k$, consider that:

$$\frac12 + \frac13 + \frac14 + \frac15 + \frac16 + \frac17 + \frac18 + \frac19 + \frac1{10} + \frac1{11} + \frac1{12} + ...$$

...is certainly a larger sum than:

$$\frac12 + \frac14 + \frac14 + \frac18 + \frac18 + \frac18 + \frac18 + \frac1{16} + \frac1{16} + \frac1{16} + \frac1{16} + ...$$


$$\frac12 + \frac12 + \frac12 + ...$$

So think of it this way. No matter how many $\frac12$'s you want to add together, you'll eventually get there. If you want $n \frac12$'s to add together, you will get there by (around) the ($2^n$)'th term. So any number you come up with can be topped by going further into the series, which means the series diverges to infinity: you can't place any finite upper bound on it that won't be surpassed.

For $ ± \frac1k$ with the $-1$'s alternating, again we look at the partially expanded series:

$$1 - \frac12 + \frac13 - \frac14 + \frac15 - \frac16 + \frac17 - \frac18 + \frac19 - \frac1{10} + \frac1{11} - \frac1{12} + ...$$

If we combine each pair of terms, we get the series:

$$\frac12 + \frac1{12} + \frac1{30} + \frac1{56} + ...$$

This is clearly bounded below because every term is positive. To show convergence we have to find some finite upper limit this can't ever exceed. Doing this directly is tricky, but we can reduce it to a common problem we already know the answer to. The series is equivalent to:

$$\frac1{1*2} + \frac1{3*4} + \frac1{5*6} + \frac1{7*8} + ...$$

...which is obviously less than:

$$\frac1{1^2} + \frac1{2^2} + \frac1{3^2} + \frac1{4^2} + ...$$

...which, you probably already know, converges to a finite value.

The moral of all this writing I'm doing is that you should try some basic mathematical logic first to see where it can take you. None of what I did above required any sort of formula (with the exception of the very last step which relies on a common proof you can look up). It just required me to start expanding the series, group terms, and see what can happen. You have to be careful when doing this with an infinite series, but the logic holds for the above examples.

The best part about being able to do this logically as opposed to formulaically is that logic can be applied to any problem of this sort, while formulas don't always work or cover what you need--and plus, why use a formula if you don't understand why that formula works in the first place? This is common in computer science in the analyzation of recursive running times: using the "master method" formula won't work in all cases, but expanding out the series and using logic will always work.

  • 1
    $\begingroup$ For information about typesetting mathematics here, please refer to this page. $\endgroup$
    – user642796
    Jul 29, 2014 at 5:43
  • $\begingroup$ @ArthurFischer Thank you! $\endgroup$ Jul 29, 2014 at 11:34
  • $\begingroup$ I believe it is fixed, but hopefully I will get feedback in the future in the event I'm not formatting the math correctly. Thanks again for the link. $\endgroup$ Jul 29, 2014 at 12:16

OP wants a single test that can distinguish convergence and divergence of the 2 aforementioned series. So far all the answers used Cauchy's convergence test for first series and are essentially correct, albeit the presentations are questionable.

Actually Cauchy's test can also be applied on second series with some extra handling. Please refer to Wikipedia article for the variables used here. We need to prove for arbitrary positive real number $\epsilon$ there exists natural number $N$ so that for any $n \gt N$ and $p \ge 1$, $$-\epsilon < \sum\limits_{n+1}^{n+p} \frac{(-1)^{k+1}}{k} < \epsilon$$

We pick $N$ so that $\frac{1}{N}$ is largest value smaller than $\epsilon$, i.e. $N=\lceil\frac{1}{\epsilon}\rceil$ (bracket means ceiling). Here 4 cases will need to be presented depending on evenness of $n$ and $p$.

1. $n$ even, $p$ even: (implying $n+1$ is odd and $n+p$ is even)

With different grouping on the terms, $$\sum\nolimits = (\frac{1}{n+1}-\frac{1}{n+2}) + {...} + (\frac{1}{n+p-1}-\frac{1}{n+p}) > 0 > -\epsilon$$ $$\sum\nolimits = \frac{1}{n+1} - (\frac{1}{n+2}-\frac{1}{n+3}) - {...} -\frac{1}{n+p} < \frac{1}{n+1} < \frac{1}{N} < \epsilon$$

2. $n$ even, $p$ odd: (implying both $n+1$ and $n+p$ are odd)

$$\sum\nolimits = (\frac{1}{n+1}-\frac{1}{n+2}) + {...} + (\frac{1}{n+p-2}-\frac{1}{n+p-1})+\frac{1}{n+p} > 0 > -\epsilon$$ $$\sum\nolimits = \frac{1}{n+1} - (\frac{1}{n+2}-\frac{1}{n+3}) - {...} - (\frac{1}{n+p-1}-\frac{1}{n+p}) < \frac{1}{n+1} < \frac{1}{N} < \epsilon$$

Other cases are left as exercise to OP, the presentation is becoming a bit tedious here.


We do not answer the specific question, but instead look at two related series, $$1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\cdots\tag{1}$$ and $$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{8}+\frac{1}{8}-\frac{1}{8}+\frac{1}{8}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\cdots\tag{2}$$

Series (1) diverges. For the sum of the first $3$ terms is $2$, the sum of the first $7$ is $3$, the sum of the first $15$ is $4$, the sum of the first $31$ is $5$, and so on. The partial sums "blow up," albeit with a great deal of reluctance. It takes an awful lot of terms to get a sum of $100$.

Series (2) converges. The partial sums are $1, \frac{1}{2}, 1, \frac{3}{4}, 1, \frac{3}{4}, 1,\frac{7}{8}, 1,\frac{7}{8}$, and so on. It is clear that the partial sums approach $1$.


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