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

Question

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 saying, for a sequence $x_n$ to converge, it is not necessary that $x_{n+1} < x_n$, for eg. $\frac{(-1)^n}{n}$. Maybe this approach works here, i.e. having $x_{n+1} < x_n$ does imply convergence, but the converse is not true. — stoic-santiago, Oct 29 '20 at 02:31
Does this answer your question? Proof of $\lim_{n\to \infty} \sqrt[n]{n}=1$ — DeBARtha, Oct 29 '20 at 02:35

score 4 · Accepted Answer · answered Oct 29 '20 at 03:03

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?

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

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