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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

Mathematics Asked on December 18, 2021

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?

2 Answers

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.

Answered by Rahul Shah on December 18, 2021

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.

Answered by Angelo on December 18, 2021

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