Does $\sum \frac{2^n}{a_n}$ converge?

Question

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.

For $n$ big enough, $0<\frac{n}{\log(a_n)}<1$. Hence the conclusion. — Paresseux Nguyen, Jun 21 '22 at 23:41
@koro Thats obvious due to the convergence of $\sum 1/log(a_n)$ — aaa bbb, Jun 21 '22 at 23:48
@sku i belive you ment $\frac {2^n}{e^{n^{1.1}}}$. To make things more simple we can take 2 instead of e and $\frac {2^n}{2^{n^{1.1}}}$ should converge. — aaa bbb, Jun 21 '22 at 23:59
@aaabbb: First, you can note that $a_n>1$ for $n$ big enough. Then, by the monotonicity of $(a_n)$, you imply that for $n$ big enough $ \frac{n}{ \log( a_{2n})} \le \sum_{m=n+1}^{2n} \frac{1}{ \log(a_m)}$. Finally, note that $\lim_{n \rightarrow \infty} \sum_{m=n+1}^{2n} \frac{1}{ \log(a_m)} =0$. — Paresseux Nguyen, Jun 22 '22 at 00:23
Use the estimate $a_n> e^{2n}$ from https://math.stackexchange.com/a/4476776. — Martin R, Jun 22 '22 at 08:28
That's an immediate consequence of the Cauchy condesation test (https://en.wikipedia.org/wiki/Cauchy_condensation_test), with $f(n)=1/\log(a_n)$. — wasn't me, Jun 23 '22 at 06:00

Max Demirdilek · Answer 1 · 2023-02-21T08:47:43.893

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.

Does $\sum \frac{2^n}{a_n}$ converge?

1 Answers1