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I am trying to determine whether the following statement is true or false.

Let $a_n$ be an unbounded, non decreasing sequence s.t. $\sum 1/\log(a_n)$ converges.

Then the series $\sum \frac{2^n}{a_n}$ conerverges.

I tried to prove the statement using the comparison test, and I get the expression $\frac{2^n \log(a_n)}{a_n}$.

Im not so sure how to go further with the calculations.

Any hints will be very appreciated.

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  • $\begingroup$ For $n$ big enough, $0<\frac{n}{\log(a_n)}<1$. Hence the conclusion. $\endgroup$
    – Paresseux Nguyen
    Commented Jun 21, 2022 at 23:41
  • $\begingroup$ @ParesseuxNguyen Can you explain further? $\endgroup$
    – aaa bbb
    Commented Jun 21, 2022 at 23:45
  • $\begingroup$ Note that $\sum \frac 1{a_n}$ converges. $\endgroup$
    – Koro
    Commented Jun 21, 2022 at 23:45
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    $\begingroup$ Related. $\endgroup$
    – metamorphy
    Commented Jun 22, 2022 at 2:02
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    $\begingroup$ That's an immediate consequence of the Cauchy condesation test (en.wikipedia.org/wiki/Cauchy_condensation_test), with $f(n)=1/\log(a_n)$. $\endgroup$
    – wasn't me
    Commented Jun 23, 2022 at 6:00

1 Answer 1

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The series $\sum\frac{2^n}{a_n}$ converges.

As shown in Ryszard Swarc's answer here, there exists an $N$ such that $a_n>e^{2n}$ for all $n>N$. For such an $N$ we then have $$\sum_{k=N+1}^{\infty}\frac{2^n}{a_n}\leq \sum_{k=N+1}^{\infty}\frac{2^n}{e^{2n}}= \sum_{k=N+1}^{\infty}\Big(\frac{2}{e^2}\Big)^n.$$

The last series is a geometric series, which converges since $\frac{2}{e^2}<1$. The claim then follows from the comparison test.

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