I am given the series $$\sum_{n=1}^{\infty}\frac{n^{n-1}}{\left ( 2n^2+n+1 \right )^{\left (n+\frac{1}{2} \right )}}.$$
Does this series converge? Solution states that comparison test should be used, though it is not stated which one. The one which looks the most promising to me is the one that says that if for two sequences $a_n \sim b_n$ then $\sum_{n=1}^\infty a_n$ is equiconvergent to $\sum_{n=1}^\infty b_n$. This requires me to find the sequence $b_n$. So I went ahead, done some gymnastics and found $b_n = \left(\frac{1}{2n} \right)^n$, meaning that my original series would be convergent. I am wondering is this valid, that is whether I found $b_n$ correctly, or not. They never really explained the process of finding $b_n$ so I am not sure which "transformation" am I allowed to do. I hope that someone can confirm my solution, if it is correct, or if it is not, provide an answer that would use one of the comparison tests to determine the convergence of this series. I have not tried other methods, since the comparing sequences is an expected way to solve it, but I think that this could also be done using the root test.
For completeness, here is my work for finding $b_n$. The reasoning I used is that I can remove terms of lower exponents everywhere, and that would still preserve asymptotic equivalence. Again, I am not sure if this is true, as it was never explained properly during the course. I just believe that that is how the process works. I also graphed both series in Desmos and it seemed that they started meeting at some point along the x axis and then continuing to converge towards $0$ together. Anyways, this is what I did:
$$a_n = \frac{n^{n-1}}{\left ( 2n^2+n+1 \right )^{\left (n+\frac{1}{n} \right )}} \sim \frac{n^{n}}{\left ( 2n^2+n+1 \right )^n} \sim \frac{n^{n}}{( 2n^2)^n} \sim \frac{n^{n}}{ 2^n n^{2n}} \sim \frac{1}{ 2^n n^n} \sim \frac{1}{(2n)^n} = b_n. $$
Maybe this solution is correct. I would still very much appreciate if someone could write an answer explaining in what ways I can manipulate the original series in order to preserve the relation of asymptotic equivalence. I have no idea if what I did is right, or why is it right if it is, or why it would not be if it is not.
Best Answer
$$\frac{n^{n-1}}{\left ( 2n^2+n+1 \right )^{\left (n+\frac{1}{n} \right )}}=\frac1{n\sqrt[n]{2n^2+n+1}}\cdot\left(\frac n{2n^2+n+1}\right)^n$$
Observe that both factors have zero as limit, thus:
$$\frac1{n\sqrt[n]{2n^2+n+1}}\cdot\left(\frac n{2n^2+n+1}\right)^n\le\left(\frac n{2n^2+n+1}\right)^n\le \left(\frac12\right)^n$$
and the comparison test gives us the answer.