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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.

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  • $\begingroup$ $\ln(n+1)\approx\ln(n), \ln(an+1)\approx\ln(an)$ $\implies$ $\frac{n^{\ln(n+1)}}{n^{\ln(an+1)}}\approx\frac{n^{\ln(n)}}{n^{\ln(an)}}=\frac{n^{\ln(n)}}{n^{{\ln(a)}+\ln(n)}}=\frac{n^{\ln(n)}}{n^{\ln(n)}\cdot n^{\ln(a)}}$ $\endgroup$ – coreyman317 Feb 3 at 15:41
  • $\begingroup$ @coreyman317 Typo: $\ln (an)=\ln(a)+ \ln(n)\neq \ln(a)\times \ln(n)$. $\endgroup$ – lulu Feb 3 at 15:42
  • $\begingroup$ Thanks, fixed! @lulu $\endgroup$ – coreyman317 Feb 3 at 15:43
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    $\begingroup$ I would also note that the summand can be written as $\frac{1}{n^{\ln \frac{an +1}{n+1}}}$ $\endgroup$ – Ryan Goulden Feb 3 at 15:49
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$$\log(n+1)-\log(an+1)=\log\left(1+\dfrac1n\right)-\log\left(1+\dfrac1{an}\right)-\log a \\=\left(1-\frac1a\right)\frac1n+o\left(\dfrac1n\right)-\log a.$$

Then the general term is asymptotic to $n^{-\log a}$ (because $\sqrt[n]n$ tends to $1$), for which the convergence condition is known.

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