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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.

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  • $\begingroup$ 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
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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.

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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 $\sum \frac1{n}$.

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    $\begingroup$ 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
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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 $$\lim_{n\to\infty}{\bigg(\frac 12\bigg)^n}=0$$

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  • $\begingroup$ 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

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