Does $\sum_{n=1}^{\infty}a_n$ converges $\implies \sum_{n=1}^{\infty}\frac{1}{a_n}$ diverges?

Question

The statement $\displaystyle \sum_{n=1}^{\infty}a_n$ converges $\implies$ $\displaystyle \sum_{n=1}^{\infty}\cfrac{1}{a_n}$ looks natural but do we have this implication? I am checking alternating series as an counter-example but could not find one yet. What can we say about the implication?

Answer 1

The convergence of $\sum a_n$ implies that $a_n\to 0$. The convergence of $\sum\frac1{a_n}$ would imply $\frac1{a_n}\to 0$.

Answer 2

If $\sum a_n $ and $\sum b_n $ are both convergent then $$\lim_{n\to+\infty}a_nb_n=0$$

but here $$a_nb_n=a_n.\frac {1}{a_n}=1.$$

Answer 3

If $\sum a_n$ and $\sum b_n$ are both positive and convergent, then $\sum (a_n+b_n)$ must be convergent, but: $$a_n+b_n=a_n+\frac{1}{a_n}\ge 2.$$

Answer 4

The $n$-th term divergence test states that if $\lim_{n\to\infty} a_n \ne 0$, then $\sum_{n=1}^{\infty} a_n$ diverges. The contraposition of this tells us that if $\sum_{n=1}^{\infty} a_n$ converges, then $\lim_{n\to\infty} a_n = 0$. If we then apply the $n$-th term divergence test to $\frac{1}{a_n}$, we get $$\lim_{n\to\infty} \frac{1}{a_n} = \infty \ne 0$$ So, by the $n$-th term divergence test, $\sum_{n=1}^{\infty} \frac{1}{a_n}$ diverges.