Asymptotic equivalence while considering series convergence 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.
 A: $$\frac{n^{n-1}}{\left ( 2n^2+n+1 \right )^{\left (n+\frac{1}{n} \right )}}=\frac1{n\sqrt[n]{2n^2+n+1}}\cdot\left(\frac n{2n^2+n+1}\right)^n$$
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.
A: With the help of Gary I think I've got the answer that he says I should have gotten. I am unsure if my notation is correct so I am posting it here for review .
$$a_n = \frac{n^{n-1}}{\left (2n^2+n+1 \right)^{\left( n + \frac{1}{2}\right)}} = \frac{n^n}{\left (2n^2+n+1 \right)^{n}}\cdot\frac{1}{n\sqrt{2n^2+n+1}} =\\= \frac{n^n}{ 2^n\cdot n^{2n}\left(1+\frac{1}{2n}+\frac{1}{2n^2} \right)^{\frac{2n^2}{n+1} \frac{n+1}{2n}}}\cdot\frac{1}{n\sqrt{2n^2+n+1}} \sim \\ \sim \frac{e^{-\frac{n+1}{2n}}}{2^n\cdot n^{n+1} \cdot \sqrt{2}n\sqrt{1+\frac{1}{2n}+\frac{1}{2n^2}}} \sim \frac{e^{-\frac{1}{2}}}{2^{n+\frac{1}{2}}\cdot n^{n+2}} = \frac{1}{\sqrt{2e}}\cdot\frac{1}{2^{n}\cdot n^{n+2}}$$
Writing this it seems that there was an error in my question. Namely there should be an $\frac{1}{2}$ in the exponent of the denominator instead of $\frac{1}{n}$. The derivation above assumes the corrected version and the question is edited now.
A: The ratio test works fine
$$a_n=
\frac{n^{n-1}}{\left ( 2n^2+n+1 \right )^{\left (n+\frac{1}{2} \right )}}$$ Take logarithms
$$\log(a_n)=(n-1)\log(n)-\left (n+\frac{1}{2} \right)\log\left ( 2n^2+n+1 \right)$$
Using Taylor for large values of $n$
$$\log(a_n)=n \log \left(\frac{1}{2 n}\right)+\frac{1}{2} \left(\log \left(\frac{1}{2
   n^4}\right)-1\right)-\frac{5}{8 n}+O\left(\frac{1}{n^2}\right)$$ Apply it twice and continue with Taylor series
$$\log(a_{n+1})-\log(a_n)=\left(\log \left(\frac{1}{2 n}\right)-1\right)-\frac{5}{2 n}+O\left(\frac{1}{n^2}\right)$$
$$\frac{a_{n+1}}{a_n}=e^{\log(a_{n+1})-\log(a_n)}=\frac{1}{2 e n} \left(1-\frac{5}{2 n}+O\left(\frac{1}{n^2}\right) \right)$$ Now, using the expansion of $\log(a_n)$ to $O\left(\frac{1}{n}\right)$, we then have your formula
$$a_n \sim \frac{1}{\sqrt{2e}}\,\frac{1}{2^{n}\, n^{n+2}}$$ which is an overestimate of the true $a_n$.
A: An alternative to show convergence of the given series:
Let $x_n=\frac{n^{n-1}}{\left ( 2n^2+n+1 \right )^{\left (n+\frac{1}{2} \right )}}$
$x_n^{\frac 1n}=\frac {n}{n^\frac 1n (2n^2+n+1) (2n^2+n+1)^\frac 1{2n}}\leq \frac {n}{n^\frac 1n (2n^2+n+1)}\implies \limsup x_n^\frac 1n=0<1$ and hence the result follows by root test.
A: Firstly, it is true that, if $\forall n\in \Bbb N:a_n,b_n>0$, $a_n=o(b_n)$ and $\sum b_n<\infty$, then $\sum a_n<\infty$: The asymptotic relation yields $\lim_{n\to \infty}\frac{a_n}{b_n}=0$, so $\exists n_0\in \Bbb N\forall n\in \Bbb N:n\geq n_0\implies a_n<b_n$ and by comparison criterion we get the desired convergence.
Here we have $0<a_n:=\frac {n^{n-1}}{(2n^2+n+1)^{n+\frac 12}}=\frac{n^{n-1}}{n^{2n+1}}\frac{1}{(2+\frac 1n+\frac {1}{n^2})^{n+\frac 12}}=n^{-n}\frac{1}{(2+\frac 1n+\frac {1}{n^2})^{n+\frac 12}}$, so $\frac {a_n}{n^{-n}}\to 0$ and $\sum a_n<\infty$, because $\sum n^{-n}<\infty$.
If we wanted to use asymptotic equivalence, we would observe that $a_n\sim \frac {n^{-n}}{2^{n+\frac 12}}:=b_n$ and of course $\sum b_n<\infty$, so $\sum a_n<\infty$. The last conclusion derived from the proposition $\forall n\in \Bbb N:a_n,b_n>0$, $a_n\sim b_n$ and $\sum b_n<\infty$, then $\sum a_n<\infty$: The asymptotic equivalence yields $\lim_{n\to \infty}\frac{a_n}{b_n}=1$, so $\exists n_0\in \Bbb N\forall n\in \Bbb N:n\geq n_0\implies a_n<\frac 32 b_n$ and by comparison criterion we get the desired convergence again.
