How can I prove the following sequence converges Given the following n-th term sequence:
$$a_{n} = \sqrt[n]{1^2+2^2+...+n^2}$$
You're asked to evaluate the limit of the given sequence, justifying your operations. 
What strategy should I take on this? I have considered taking some inequality in order to, eventually, be able to use the Squeeze Theorem.
I've tried exploring some initial terms, viz:
$$
a_{1} = \sqrt[n]{1^2} = 1^\frac{1}{n}\\
a_{2} = \sqrt[n]{1^2 + 2^2} = 5^\frac{1}{n}\\
a_{3} = \sqrt[n]{1^2 + 2^2 + 3^2} = 14^\frac{1}{n}\\
\vdots\\ 
a_{n} = \sqrt[n]{1^2+2^2+...+n^2} = \sqrt[n]{k + n^2} = (k + n^2)^\frac{1}{n}
$$
Supposing $k$ is the sum of all the $n-1$ terms of the sequence, rightly before $n²$. We can see that $a_{n}$ is always smaller than $a_{n+1}$ for any $n$ strictly positive. 
I'm not sure, though, what else I can try. I would appreciate a hand here. 
Answer is:

 \begin{align}  1 \end{align}


$$1 = \sqrt[n]{1^2} \leq \sqrt[n]{1^2 + 2^2 + \ldots + n^2} \leq \sqrt[n]{n^2 + n^2 + \ldots + n^2} = \sqrt[n]{n^3} \xrightarrow{n\to\infty}1$$
By the squeeze theorem, the sequence converges to $1$.
