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, $\lim \sqrt[n]{1^2+2^2+...+n^2} =1$.

Note that $\sum_{k=1}^{n} k^a \sim  \frac{n^{a+1}}{a+1}$
because $$\lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^{n} (k/n)^a =\int_{0}^{1} x^a dx=\frac{1}{a+1}$$
So $$L=\lim_{n \to \infty} \left [\sum_{k=1}^{n} k^2 \right]^{1/n}=\lim_{n \to \infty} (n^3/3)^{1/n}= \lim_{n\to \infty} \exp[\frac{1}{n} \ln (n^3/3)]=\lim_{n \to \infty} \exp[\frac{1}{n}(3 \ln n-\ln3)]=e^{0}=1. $$
