I have been given the following question:
Let $ (x_n)_{n \in \mathbb{N}} $ be a sequence of real, positive numbers such that $ x_n \rightarrow L $ as $ n \rightarrow \infty $. If $L>0$, prove that the following series converges:
$$\displaystyle\sum_{n=1}^\infty\frac{n^2}{n^4x_n+1}$$
I'm very much at a loss as to how to proceed with this...
I had initially went down the path of saying $$ \frac{n^2}{n^4x_n+1} < \frac{n^2}{n^4x_n}=\frac{1}{n^2x_n} = \frac{1}{n^2} \cdot \frac{1}{x_n} $$
and argued that this sequence converges to $0$, however then remembered that convergence to $0$ does not implicate the convergence of the series...
Then considered using the Ratio Test, but I am a bit unsure how to apply this for the specific question. Any tips would be massively appreciated!!
The usual way to prove that a series converges is to find an asymptotic equivalent ($\sim$, $O$, ...) of the general term of the series. Here since $L>0$ $$ \frac{n^2}{n^4 x_n+1} \sim \frac{n^2}{n^4L}=\frac{1}{n^2L}. $$ and using the fact that $\sum \frac{1}{n^2}$ converges, we have the result. Another way: for large $n$ $$ 0 \le \frac{n^2}{n^4x_n+1} \le \frac{1}{n^2 x_n} \le \frac{1}{n^2(L/2)}. $$ (indeed $x_n \to L$ so taking $\epsilon=L/2 >0$ there exists $n_0$ such that $\forall n \ge n_0, x_n \ge L/2$.) And using the fact that $\sum \frac{1}{n^2}$ converges, we have the result.