Converging or diverging series.

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.

• Do you know how to apply the integral test? – Mark Viola Nov 5 '18 at 0:27
• yes , however the series shouldn't the series be decreasing and positive? isn't it increasing for $n <4$? – 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 $$\sum \frac1{n}$$.

• How is $\log\left(\frac{n+1}{2n+1}\right)\sim \frac12$? – Mark Viola Nov 5 '18 at 1:28
• @MarkViola opssss...I lost a log! Thanks :) – 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 $$\lim_{n\to\infty}{\bigg(\frac 12\bigg)^n}=0$$

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