Let's say I have $u_n = f(n) > 0$, such that $f(n) \sim g(n)$ and such that there exist $\alpha \geq 1$ such that : $g(n) = o(1/n^\alpha)$, then can I say that the series of general term : $u_n$ converges ?
For example if : $u_n = \frac{n^4\ln(3n)}{e^{2n}}$ then can I say the following :
$u_n \sim \frac{n^4\ln(n)}{e^{2n}}$ and because $\frac{n^4\ln(n)}{e^{2n}} = o(1/n)$ then $\sum u_n$ converges ?
Thank you,
On
Yes we can say that the series converges by limit comparison test with $\sum \frac 1 {n^a}$ for $a>1$ since
$$\large{\frac{\frac{n^4\ln(n)}{e^{2n}}}{\frac 1 {n^a}}\to 0}$$
Note that in general by limit comparison test it suffices that $u_n \sim \frac1 {n^a}$ with $a>1$ for convergence but it is not necessary as for example for $u_n=\frac1{n\log^2 n}$.
For your general question, the Bertrand's series: $$\sum_{n\ge 2}\frac1{n\log n} $$ is a counter-example (by the integral test). If you have $\alpha>1$, you can answer positively.