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Prove or disprove: $$\sum_{k=0}^{\infty}(-1)^{k}\frac{1}{2k+1}$$ is convergent.

I have started trying with ratio test but at the end I had $1$ as result which is bad :D

(Would you even recommend me trying ratio test on an alernating series?)

Then I have used another theorem but I don't know its name. Show that it's a zero-sequence and then show it's monotonic decreasing:

$$\lim_{k\rightarrow\infty}\frac{1}{2k+1}=0$$

Now show it's monotonic decreasing:

$$\frac{1}{2k+1}\geq \frac{1}{2(k+1)+1}$$

$$\frac{1}{2k+1} \geq \frac{1}{2k+3}$$

$$1 \geq \frac{2k+1}{2k+3}$$

Thus it's monotonic decreasing. So all in all, the series converges.


Did I do it correctly?

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  • $\begingroup$ hint: alternating series test $\endgroup$
    – tired
    Commented Sep 19, 2016 at 16:14
  • $\begingroup$ The absolute value of the terms decreases to $0$, so it converges. Note: if the terms did not decrease to $0$ (but still went to $0$) that would not prove divergence. It would just mean that you couldn't apply the alternating series test. As it happens, though, they do decrease to $0$. $\endgroup$
    – lulu
    Commented Sep 19, 2016 at 16:15
  • $\begingroup$ Yes, you're using the Alternating Series Test correctly. $\endgroup$
    – user84413
    Commented Sep 19, 2016 at 16:15
  • $\begingroup$ Oh alright, that's what I have used ^^ $\endgroup$
    – tenepolis
    Commented Sep 19, 2016 at 16:15
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    $\begingroup$ Correct. The Leibniz test is great...it shows that alternating series wherein the terms decrease to $0$ converge. the converse, however, is false. Worth noting: it's pretty hard to work with general series that converge but only conditionally. $\endgroup$
    – lulu
    Commented Sep 19, 2016 at 16:19

1 Answer 1

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In the end you got a true statement: $1 \le 3$, which means the original equivalent statement, $\frac{1}{2k+1} \ge \frac{1}{2(k+1) + 1}$ is also true, so it is monotonically decreasing (edit: you corrected this).

The theorem you are referring to is the alternating series test. It says that an alternating series converges if the absolute value of the terms decreases monotonically to zero. Thus, since you have shown that it decreases, and we can see that the series has $(-1)^k$ so it is alternating, your reasoning is correct: the series converges. In general if you want to prove an alternating series converges, alternating series test is a good way to go.

However, an important word of warning: if you have an alternating series where the terms do not monotonically decrease, that does not mean it diverges. That makes the alternating series test inconclusive. In an earlier draft of your question, you seemed to think that if it was not monotonically decreasing, it must diverge -- but this is not true.

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    $\begingroup$ Wow I see you wrote that answer very fast so you solved the task very fast and explain well same time, I want be that fast too.. ^^ Thank you very much! $\endgroup$
    – tenepolis
    Commented Sep 19, 2016 at 16:20
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    $\begingroup$ @tenepolis You're very welcome! $\endgroup$ Commented Sep 19, 2016 at 16:21

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