If $\sum_{n=1}^{\infty} a_n$ converges, then $\lim_{n \to \infty} a_n = 0$. Therefore, if $\lim_{n \to \infty} a_n$ does not exist, or exists but is not zero, then the series $\sum_{n=1}^{\infty} a_n$ is divergent.

Now, I think the whole point of such a theorem is to use the limit to test if a series converges or diverges. But apparently, the converse of this theorem is false. We cannot conclude that if $\lim_{n \to \infty} a_n = 0$, then $\sum_{n=1}^{\infty} a_n$ converges. For example, the harmonic series is a counter example to this.

But I guess my question then is: why is there such a theorem if it doesn't go in both ways? I mean, if I know that my series converges, then I automatically know that the limit of the series will be zero. That's not really interesting. I would want the limit to give me information about my series, but that's not possible here, because if the $n$th term approaches $0$ as $n$ approaches $\infty$, then the series may or may not converge.

  • $\begingroup$ Knowing that a series is divergent is useful information. As you said, you take the limit, it doesn't exist or is not zero them you know the series is divergent. $\endgroup$ – Pp.. Dec 31 '14 at 14:05
  • $\begingroup$ You are correct in noticing that if the $n$th term approaches 0, then you cannot immediately tell if the series converges. What you *can* tell is the **contrapositive** which states that if the terms $a_n$ do not approach zero, then the series cannot converge. Example: $\sum_{n=1}^\infty (-1)^n$. Why we care about the theorem despite it not being a necessary and sufficient condition is that it is generally incredibly easy to check for the necessary condition and can save a great deal of time. $\endgroup$ – JMoravitz Dec 31 '14 at 14:06
  • $\begingroup$ Yes, it is useful if a series diverges. But it doesn't tell us anything about whether it converges or not. $\endgroup$ – Kamil Dec 31 '14 at 14:10

There are multiple convergence tests that can check if a series converges. The "holy grail" would be a test that is easy, works on every sequence, and is never inconclusive (it always says if the series converges or does not converge).

Unfortunately, there is no such test. The main tests, the ratio, root, and the currently discussed test (limit of the summand), all have inconclusive cases. In our current test, if the limit of the summand is zero, the test is inconclusive. If the ratio of consecutive terms or $n$'th root of the $n$'th term tend to one those tests are inconclusive.

Some tests, like the integral test, have no inconclusive case. However, they cannot reasonably be applied to every series. Too many integrals are not known! To use the comparison tests you need another series that is already known and relevant. And so on.

There is no perfect convergence test. My first-year analysis class had a devious series of homework problems designed to press that idea into our brains. Look at the good side: there is job security for mathematicians!

To sum up: This test works often enough to be useful. That is why it is used, even if it does not always help. It is one more tool in your toolbox.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.