In my practice midterm there is a multiple choice question that I thought was relatively straight forward but the solutions gave an answer that was unexpected to me.
Question: If $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ are convergent series, which of the following is not necessarily true?
(A)$\sum_{n=1}^\infty a_nb_n$ is convergent
(B)$\sum_{n=1}^\infty (a_n+b_n)$ = ($\sum_{n=1}^\infty a_n$) + ($\sum_{n=1}^\infty b_n$)
(C)$\sum_{n=1}^\infty (a_n-b_n)$ = ($\sum_{n=1}^\infty a_n$) – ($\sum_{n=1}^\infty b_n$)
(D)$\sum_{n=1}^\infty ca_n$=$c\sum_{n=1}^\infty a_n$ for any constant c
(E)$\sum_{n=1}^\infty \frac{1}{a_n}$ is divergent (assuming $a_n\ne0$ for all n)
I understand why options B, C & D are true given the Algebraic Properties of Convergent Series and I thought that A is true as well. However, the solutions say that the correct answer is A.
Is there any proof that holds E to be true and under what situations would A be false in this scenario?
Best Answer
If $\sum_\limits{n=1}^\infty a_n$ is convergent then there exists $\lim_\limits{n\to\infty} a_n=0$, hence $\lim_\limits{n\to\infty}\frac{1}{a_n}=\infty$ and $\sum_\limits{n=1}^\infty\frac{1}{a_n}$ cannot be convergent (assuming $a_n\ne0$ for any $n\in\mathbb{N}$).
So (E) is necessarily true.
But (A) is not necessarily true, indeed $\sum_\limits{n=1}^\infty a_n=\sum_\limits{n=1}^\infty(-1)^n\frac{1}{\sqrt[3]{n}}\;$ and $\;\sum_\limits{n=1}^\infty b_n=(-1)^n\frac{1}{\sqrt[6]{n}}\;$ are convergent, but $\sum_\limits{n=1}^\infty a_nb_n=\sum_\limits{n=1}^\infty\frac{1}{\sqrt{n}}$ is divergent.