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

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?

-
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

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).

-

Hint.

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

-

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.

-

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.

-