Convergence of series with logarithm upon polynomial as terms [duplicate]

Solution 1:

For $x>1$, let $x=1+2a, a>0$, then $$\lim_{n\to\infty}\frac{\frac{\log n}{n^{1+2a}}}{\frac{1}{n^{1+a}}}=\lim_{n\to\infty}\frac{\log n}{n^{a}}=0.$$

Because the series $\sum\frac{1}{n^{a+1}}$ converges, so does $\sum \frac{\log n}{n^{x}}.$

Solution 2:

The reason you can't do that is because the limit of those terms is $1,$ so the ratio test won't work here. Instead, note that $\displaystyle f(t) = \frac{ \log t}{t^x} $ is eventually decreasing, since $\displaystyle f'(t) = \frac{1- x \log t}{t^{x+1}}$. Then either the integral test or Cauchy condensation will finish this off.