Determine value of the limit:
$$L=\lim_{n\to \infty}\sqrt[n]{1^2+2^2+\cdots+n^2}$$
My try:
$$1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$$
Hence: $$L=\lim_{n\to \infty} \sqrt[n]{\frac{n(n+1)(2n+1)}{6}}=?$$
But, come here, i do not know how, because please help me.
$$1\leftarrow\sqrt[n]{n}<\sqrt[n]{1^2+\cdots+n^2}<\sqrt[n]{n^3}\to 1$$
The limit used above can be shown via AM-GM for instance:
$$1+\frac{2}{\sqrt{n}}-\frac{2}{n}=\frac{1}{n}\Big(1+\cdots+ 1+\sqrt{n}+\sqrt{n}\Big)\geq \sqrt[n]{n}\geq 1$$
Since $n^3 < n(n+1)(2n+1) < n*2n*3n = 6n^3$, you have $$n^{3 \over n} < (n(n+1)(2n+1))^{1 \over n} < 6^{1 \over n} n^{3 \over n}$$ Since $6^{1 \over n}$ goes to $1$, in order to show your limit is $1$ it suffices to show that $$\lim_{n \rightarrow \infty} n^{3 \over n} = 1$$ You can do this via L'hopital 's rule for example.
