Prove $sum_{n=1}^infty frac{1}{a_n}$ is divergent if $sum_{n=1}^infty a_n$ and $sum_{n=1}^infty b_n$ are both convergent

Question

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)$sum_{n=1}^infty ca_n$=$csum_{n=1}^infty a_n$ for any constant c
(E)$sum_{n=1}^infty frac{1}{a_n}$ is divergent (assuming $a_nne0$ 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?

Rahul Shah · Answer

Talking about E, if a series converges then the corresponding sequence must converge to $0$ . So clearly it's reciprocal will not converge to 0 ( infact it will diverge to  infinity).
Now for A, if you consider an and bn both to be $dfrac {(-1)^n}{sqrt n}$. They both converge by alternating series test. but the product is $1/n$ which clearly diverges. You can very easily check that this sequence satisfies the required conditions of the alternating series test.
Ps : if you consider two absolutely convergent series then their product does indeed converge.

Angelo · Answer

If $sum_limits{n=1}^infty a_n$ is convergent then there exists $lim_limits{ntoinfty} a_n=0$, hence $lim_limits{ntoinfty}frac{1}{a_n}=infty$ and $sum_limits{n=1}^inftyfrac{1}{a_n}$ cannot be convergent (assuming $a_nne0$ for any $ninmathbb{N}$).
So (E) is necessarily true.
But (A) is not necessarily true, indeed $sum_limits{n=1}^infty a_n=sum_limits{n=1}^infty(-1)^nfrac{1}{sqrt[3]{n}};$ and $;sum_limits{n=1}^infty b_n=(-1)^nfrac{1}{sqrt[6]{n}};$ are convergent, but $sum_limits{n=1}^infty a_nb_n=sum_limits{n=1}^inftyfrac{1}{sqrt{n}}$ is divergent.

Prove $sum_{n=1}^infty frac{1}{a_n}$ is divergent if $sum_{n=1}^infty a_n$ and $sum_{n=1}^infty b_n$ are both convergent

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