If $\sum_{n=1}^{\infty} a_n$ converges, does $\sum_{n=1}^{\infty}\frac{1}{a_n}$ diverge?

Question

If a series $\sum_{n=1}^{\infty} a_n$ converges, does $\sum_{n=1}^{\infty} \dfrac{1}{a_n}$ diverge to infinity?

Answer 1

It depends on your definition of "diverge". Some take it as "have a $\pm\infty$ limit" other take it as "non convergent"

I'll stick with the wikipedia definition (emphasys mine)

In mathematics, a divergent series is an infinite series that is not convergent, meaning that the infinite sequence of the partial sums of the series does not have a finite limit.

If a series converges, the individual terms of the series must approach zero. Thus any series in which the individual terms do not approach zero diverges.

According to this definition, $\sum_{n}\frac{1}{a_n}$ clearly diverge if $\sum_{n}a_n$ converge. Because $a_n$ and $\frac{1}{a_n}$ can't both have $0$ as a limit.

But that doesn't mean that $\sum_{n}a_n$ as an infinite limit. The series can have no limit at all, as per @lulu example in the comments:

$$a_n = \frac{{(-1)}^n}{n}$$

Answer 2

A series converges if and only if the $\lim_{n \to \infty} a(n) = 0$. Therefore, $\lim_{n \to \infty} {1\over a(n)} = \infty$. Clearly, $\lvert a(n)\rvert < \lvert a(n+1)\rvert$. Because each term is larger (in the absolute sense) than the previous, it diverges to either $+\infty$, $-\infty$, or oscillates between both.

Answer 3

No. In fact, given that $\sum a_n$ converges, $\sum\frac{1}{a_n}\to\infty$ if and only if there is some $N \in \mathbb{N}$ such that for all $n > N$, we have $a_n > 0$ and $a_n \neq 0$ for all $n$.

Proof:

In the forwards direction, since $\sum \frac{1}{a_n}\to\infty$, there is, in particular, some $N \in\mathbb{N}$ such that for all $n > N$, we have $\frac{1}{a_n} > 0$, hence $a_n > 0$. In the reverse direction, the necessity of $a_n \neq 0$ is obvious. For any $C > 0$, it is also true that $\frac{1}{C} > 0$, and $(a_n)$ is null (since $\sum a_n$ converges), so there is some $M\in\mathbb{N}$ such that for all $n > M$, $|a_n| < \frac{1}{C}$. Thus, taking $L = \max(N,M)$, for any $n > L$, we have $0 < a_n < \frac{1}{C}$, so $\frac{1}{a_n} > C$, hence (up to a constant that you can feel free to get rid of for yourself), $\sum\frac{1}{a_n} > C$, hence the result.