A proof in $\varepsilon$-language for $\lim \sqrt[n]{1^2+2^2+...+n^2} = 1$ I found a proof that $\lim \sqrt[n]{1^2+2^2+...+n^2}=1$ by $\varepsilon$-language, but I think it's quite complicated and not sure that it's correct.
My question is:
1- Is my proof correct?
2- Is there another simpler proof in the sense of $\varepsilon$-language?
Please help me. Thanks.
Solution:
Let $a_n=\sqrt[n]{1^2+2^2+...+n^2}-1$. Then $a_n>0$ and:
$$ (a_n+1)^n= 1^2+2^2+...+n^2 = \frac{n(n+1)(2n+1)}{6}$$
By the binomial theorem we have:
$ (a_n+1)^n = 1 + na_n+\frac{n(n-1)}{2}a_n^2+\frac{n(n-1)(n-2)}{6}a_n^3+\frac{n(n-1)(n-2)(n-3)}{24}a_n^4+\cdots+a_n^n $
Since $a_n>0$, then $(a_n+1)^n>\dfrac{n(n-1)(n-2)(n-3)}{24}a_n^4$ and therefore:
$$\dfrac{n(n+1)(2n+1)}{6}>\dfrac{n(n-1)(n-2)(n-3)}{24}a_n^4$$
This is equivalent to
$$(n+1)(2n+1)>\dfrac{(n-1)(n-2)(n-3)}{4}a_n^4 \Leftrightarrow a_n^4<\dfrac{4(n+1)(2n+1)}{(n-1)(n-2)(n-3)}$$
$$\Rightarrow a_n^4< \dfrac{8(n+1)^2}{(n-2)^2(n-3)}=\dfrac{8}{n-3}\Big(1+\dfrac{3}{n-2}\Big)^2<\dfrac{128}{n-3}.$$
Thus: $a_n<\sqrt[4]{\dfrac{128}{n-3}}$. For every $\varepsilon >0$, take $N>3+\dfrac{128}{\varepsilon^4}$, then for all $n\ge N$,
$$ n-3 \ge N -3 > \dfrac{128}{\varepsilon^4} $$
Thus: $ \varepsilon^4 > \dfrac{128}{n-3} \Rightarrow \varepsilon > \sqrt[4]{\dfrac{128}{n-3}} >a_n $.
Hence, $\lim a_n=0$ or equivalently, $\lim \sqrt[n]{1^2+2^2+...+n^2} =1$.

because
\begin{gather*}
1\leq \sqrt[n]{1^2+2^2+\cdots+n^2}\leq \sqrt[n]{n\cdot n^2}
=(\sqrt[n]{n})^3,\end{gather*}
and
\begin{gather*}
\lim_{n\to\infty}1=1=\lim_{n\to\infty}(\sqrt[n]{n})^3=\left(\lim_{n\to\infty}\sqrt[n]{n}\right)^3=1,
\end{gather*}
we have, by the squeeze test,
$\lim_{n\to\infty}\sqrt[n]{1^2+2^2+\cdots+n^2}=1.$
