A proof in $\varepsilon$-language for $\lim \sqrt[n]{1^2+2^2+...+n^2} = 1$

Question

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

Answer 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, $\lim_{n\to\infty}\sqrt[n]{1^2+2^2+\cdots+n^2}=1.$

Answer 2

A computation without $\epsilon$ but assuming a closed formula for the sum of the first $n$ squares.

For $n\ge 1$, $a_n=\sqrt[n]{\frac{n(n+1)(2n+1)}{6}}=\exp\left(\frac{1}{n}\ln\left(\frac{n(n+1)(2n+1)}{6}\right)\right) \to \exp(0)=1$.

Answer 3

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 $$L=\lim_{n \to \infty} \left [\sum_{k=1}^{n} k^2 \right]^{1/n}=\lim_{n \to \infty} (n^3/3)^{1/n}= \lim_{n\to \infty} \exp[\frac{1}{n} \ln (n^3/3)]=\lim_{n \to \infty} \exp[\frac{1}{n}(3 \ln n-\ln3)]=e^{0}=1. $$