Question

\[\lim_{n\to\infty }\sqrt[n]{1^2+2^2+...+n^2}\]

Answer 1

i have a guess
maybe write the closed form of
\[\sum_{k=1}^n k^2\]

Answer 2

\[\lim_{n\to \infty}(\frac{n(n+1)(2n+1)}{6})^{\frac{1}{n}}\]might be a good start

Answer 3

then perhaps l'hopital

Answer 4

just a guess mind you, but it seem possible now
take the log etc

Answer 5

I wrote it that way.
what i am confused with is that do i need to show individually that n1/n, \((n+1)^{1/n}\) and other terms converge to 1? Basically here all terms will converge to 1.
And additionally, i haven't learnt l'hopital so far. So, i am not supposed to use that. I guess.
I was wondering if there is a better or a shortcut way.

Answer 6

\[\begin{align*}\lim_{n\to \infty}\left(\frac{n(n+1)(2n+1)}{6}\right)^{1/n}
&=\left(\lim_{n\to\infty}\frac{1}{6^{1/n}}\right)\left(\lim_{n\to \infty}n^{1/n}\right)\\&\quad\times\left(\lim_{n\to\infty}\left(n+1\right)^{1/n}\right)\left(\lim_{n\to\infty}(2n+1)^{1/n}\right)
\end{align*}\]
The first limit clearly converges to 1.

The second limit can be evaluated as follows:
\[\begin{align*}
\lim_{n\to\infty}n^{1/n}&=\lim_{n\to\infty}\exp\left(\ln n^{1/n}\right)\\\\
&=\lim_{n\to\infty}\exp\left(\frac{1}{n}\ln n\right)\\\\
&=\exp\left(\lim_{n\to\infty}\frac{\ln n}{n}\right)\end{align*}\]
which can be evaluated using the squeeze theorem (since you're l'Hopital-less).

Answer 7

Try using the same reasoning for the remaining limits.