Determine if the series $\left(\frac{\:n+3}{2n+1}\right)^{\ln(n)}$ diverges or converges.

I've tried both the root and ratio tests which got me nowhere.

I tried finding a smaller divergent series (or bigger convergent series) to apply the comparison test but failed.

Any hints or advices are much appreciated.

Thanks in advance.

  • $\begingroup$ Do you know how to apply the integral test? $\endgroup$ – Mark Viola Nov 5 '18 at 0:27
  • $\begingroup$ yes , however the series shouldn't the series be decreasing and positive? isn't it increasing for $n <4$? $\endgroup$ – Raku Nov 5 '18 at 0:30

Note that we have

$$\left(\frac{n+3}{2n+1}\right)^{\log(n)}=n^{\log\left(\frac{n+3}{2n+1}\right)}\ge n^{-\log(2)}$$

and $\log(2)<1$.

Now apply the comparison test.


We have that

$$\left(\frac{n+3}{2n+1}\right)^{\ln(n)}=e^{\ln n \cdot \ln\left(\frac{n+3}{2n+1}\right)}\sim e^{\ln n \cdot \ln \frac12}=e^{\ln n^{-\ln 2}}=\frac1{n^{\ln 2}} $$

then refer to direct comparison test or limit comparison test with $\sum \frac1{n}$.

  • 1
    $\begingroup$ How is $\log\left(\frac{n+1}{2n+1}\right)\sim \frac12$? $\endgroup$ – Mark Viola Nov 5 '18 at 1:28
  • $\begingroup$ @MarkViola opssss...I lost a log! Thanks :) $\endgroup$ – gimusi Nov 5 '18 at 6:07

As $n\to \infty, \frac{n+3}{n}\to1$ and $\frac{2n+1}{2n}\to 1$

so: $$\lim_{n\to\infty}{\frac{n+3}{2n+1}}=\lim_{n\to\infty}{\frac{n}{2n}}=\frac 12$$ Also note that $$\lim_{n\to\infty}{\ln(n)}=\infty$$ so we have $$\lim_{n\to\infty}{\bigg(\frac 12\bigg)^n}=0$$

  • $\begingroup$ I think the OP is asking for the series. $\endgroup$ – gimusi Nov 5 '18 at 0:28
  • $\begingroup$ Well, I have shown that $$\sum_{n=0}^{\infty}{\bigg(\frac{n+3}{2n+1}\bigg)^{\ln(n)}}\approx\sum_{n=0}^{\infty}{2^{-n}}$$ $\endgroup$ – Rhys Hughes Nov 5 '18 at 0:48
  • $\begingroup$ But the series diverges. $\endgroup$ – gimusi Nov 5 '18 at 6:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.