Suppose we know that $a_n \rightarrow 2$ determine whether the following series converges:
$$ \sum \frac{1}{(a_n )^{2/n}} $$
We know that since $a_n$ converges, it is bounded, suppose for some index $n_0$ we have that $1<a_n <3$, since $a_n$ must converge to $2$ eventually, this is certainly true.
We can then say that $$ \sum_{n=n_0}^\infty \frac{1}{(a_n )^{2/n}} > \sum_{n=n_0}^\infty \frac{1}{(3 )^{2/n}}=\sum_{n=n_0}^\infty \frac{1}{(9 )^{1/n}}$$ The limit of the genreal term goes as $9^{-\frac{1}{n}} \rightarrow 9^{0}=1>0$ So the minorant series diverges and the limit diverges.
Is my proof correct? Is there a quicker method to see this, perhaps?
There is a slightly shorter proof: $$\log\biggl(\frac1{(a_n)^{2/n}}\biggr)=-\underbrace{\frac2n}_{\begin{matrix}\downarrow \\ 0\end{matrix}}\,\underbrace{\log(a_n)}_{\begin{matrix}\rule{0pt}{2.3ex}\downarrow \\ \log 2\end{matrix}}$$ so the denominator tends to $1$.