Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why is test for divergence inconclusive when $\lim_{n\to\infty} a_n = 0$?

In my lecture slides, its said that

If series is convergent

$$\lim_{n\to\infty} a_n = \lim_{n\to\infty} S_n - S_{n-1} = \lim_{n\to\infty} S_n - \lim_{n\to\infty} S_{n-1} = L-L=0$$

But after that its stated that test for divergence is inconclusive when $\lim=0$. Why is that. Can't I say its convergent?

share|cite|improve this question
To paraphrase the real estate agents, three words: harmonic, harmonic, harmonic. As your responders point out so well, this is the standard example for why the test is inconclusive. – Chris Leary Feb 15 '13 at 17:19
up vote 6 down vote accepted

To perhaps make things more transparent, consider the sum $$ 1+ {1\over2}+{1\over2}+{1\over3}+{1\over 3}+{1\over3}+\cdots+ \underbrace{{1\over n}+{1\over n}+\cdots+{1\over n}}_{n\text{-terms}}+\cdots $$ The terms being added are heading towards 0, but they are not heading towards 0 fast enough to make the sum of them convergent (the sum is clearly infinite).

share|cite|improve this answer


Consider the harmonic series $\displaystyle \sum_{n=1}^\infty \dfrac 1 n$

The harmonic series diverges again and again but $\lim_{n \to \infty} \dfrac 1 n=0$.

To make things clear, the necessary condition for a series $\sum a_n$ to converge is that its $n^{th}$ term should converge to $0$, that is $a_n \to 0$. But this is not sufficient as the above example suggests.

I googled the term "Harmonic series diverges" because I was lazy to add a proof, but, not surprisingly, I was led to the links here that gives $20$ different proofs of this fact in the first link and $19$ others in the next link. I just knew one proof. sigh

share|cite|improve this answer

If they go to 0 really slowly, their sum can still be infinite. Here's an easy example, much like the harmonic series, but perhaps clearer in how slow the terms go to 0.

Let $a_1 = 1$, $a_2 = a_3 = \frac{1}{2}$, $a_4 = a_5 = a_6 = \frac{1}{3}$, $a_7 = a_8 = a_9 = a_{10} = \frac{1}{4}$, and so on. So, the first term adds to 1, the next 2 terms add to 1, the next 3 terms add to 1, the next 4 terms add to 1, the next 5 terms add to 1, the next 6 terms add to 1, ..., the next $n$ terms add to 1, ... . So, no matter how far out we, the leftover terms still add up to $\infty$.

Remember, the definition of a convergent series is that the sequence of partial sums converges. Just because the terms themselves go to 0 does not imply that the sequence of partial sums eventually converges to something. If the terms of the series go to 0 slowly enough, the partial sums will grow without bound, even though the growth might be very slow, and thus the series will diverge.

share|cite|improve this answer

Divergence test tests for divergence, not convergence. Just because a certain series fulfills the conditions for not being divergent in a divergence test does not automatically mean that the series is convergent. Thus the solution being inconclusive.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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