Prove that : $x_{n}=\sqrt[n]{n^{2}+2}$ is convergent. Prove that : $x_{n}=\sqrt[n]{n^{2}+2}$ is convergent.

I intended using inductive method. From inductive step, I supposed $x_{n}<x_{n-1}$, and I need to prove $x_{n+1}<x_{n}$. However, with $x_{n}<x_{n-1}$, I could only show that $x_{n-1} >1$. Please help me, thank you so much!

Just a hint: You can alternatively try using the Sandwich Theorem
We have, $\forall n\in\mathbb{N}$, $n^2<n^2+2\implies n^{\frac{2}{n}}<(n^2+2)^{\frac{1}{n}}$
Also, $(n^2+2)<n^3$ $\forall n\geq2\implies (n^2+2)^{\frac{1}{n}}<n^{\frac{3}{n}}$.
So, combining these two, we get,
$$n^{\frac{2}{n}}\leq(n^2+2)^{\frac{1}{n}}\leq n^{\frac{3}{n}}$$ $$\forall n\geq2$$
Can you complete this now?
