# Find nature of series

We have $$\sum_{n=1}^{\infty}\frac{n^{\ln(n+1)}}{n^{\ln(an+1)}},$$ and problem asks for nature of that series, discussed after values of parameter $$a$$. I tried with D'Alembert method but seems to become so complicated and get no result.

• $\ln(n+1)\approx\ln(n), \ln(an+1)\approx\ln(an)$ $\implies$ $\frac{n^{\ln(n+1)}}{n^{\ln(an+1)}}\approx\frac{n^{\ln(n)}}{n^{\ln(an)}}=\frac{n^{\ln(n)}}{n^{{\ln(a)}+\ln(n)}}=\frac{n^{\ln(n)}}{n^{\ln(n)}\cdot n^{\ln(a)}}$ – coreyman317 Feb 3 at 15:41
• @coreyman317 Typo: $\ln (an)=\ln(a)+ \ln(n)\neq \ln(a)\times \ln(n)$. – lulu Feb 3 at 15:42
• Thanks, fixed! @lulu – coreyman317 Feb 3 at 15:43
• I would also note that the summand can be written as $\frac{1}{n^{\ln \frac{an +1}{n+1}}}$ – Ryan Goulden Feb 3 at 15:49

$$\log(n+1)-\log(an+1)=\log\left(1+\dfrac1n\right)-\log\left(1+\dfrac1{an}\right)-\log a \\=\left(1-\frac1a\right)\frac1n+o\left(\dfrac1n\right)-\log a.$$
Then the general term is asymptotic to $$n^{-\log a}$$ (because $$\sqrt[n]n$$ tends to $$1$$), for which the convergence condition is known.