1

Given the following n-th term sequence:

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

You're asked to evaluate the limit of the given sequence, justifying your operations.

What strategy should I take on this? I have considered taking some inequality in order to, eventually, be able to use the Squeeze Theorem.

I've tried exploring some initial terms, viz: $$ a_{1} = \sqrt[n]{1^2} = 1^\frac{1}{n}\\ a_{2} = \sqrt[n]{1^2 + 2^2} = 5^\frac{1}{n}\\ a_{3} = \sqrt[n]{1^2 + 2^2 + 3^2} = 14^\frac{1}{n}\\ \vdots\\ a_{n} = \sqrt[n]{1^2+2^2+...+n^2} = \sqrt[n]{k + n^2} = (k + n^2)^\frac{1}{n} $$ Supposing $k$ is the sum of all the $n-1$ terms of the sequence, rightly before $n²$. We can see that $a_{n}$ is always smaller than $a_{n+1}$ for any $n$ strictly positive. I'm not sure, though, what else I can try. I would appreciate a hand here.

Answer is:

\begin{align} 1 \end{align}

MCarsten
  • 73
  • Hint: bound $a_n$ from above and below by converging sequences – Guy Aug 29 '17 at 18:30
  • use $1^2+2^2+\dots +n^2=\dfrac{n(n+1)(2n+1)}{6}$ and $\lim_{n\to\infty}\sqrt[n]{n}=1$ – MAN-MADE Aug 29 '17 at 18:31
  • "$a_n$ is always smaller than $a_{n+1}$"---- this is wrong. Note that $a_{n+1}=\sqrt[n+1]{a^n_n+(n+1)^2}$ does not imply that. And certainly $a_n$ strictly decreasing! – MAN-MADE Aug 29 '17 at 18:52

2 Answers2

1

Hint

Show by induction that

$$\sum_{k=1}^n k^2=\dfrac{n^3}{3}+\dfrac{n^2}{2}+\dfrac n6. $$ Thus

$$\sqrt[n]{1^2+2^2+...+n^2}\le \sqrt[n]{\dfrac{5n^3}6 }.$$

Can you conclude from here?

mfl
  • 29,399
1

$$1 = \sqrt[n]{1^2} \leq \sqrt[n]{1^2 + 2^2 + \ldots + n^2} \leq \sqrt[n]{n^2 + n^2 + \ldots + n^2} = \sqrt[n]{n^3} \xrightarrow{n\to\infty}1$$

By the squeeze theorem, the sequence converges to $1$.

mechanodroid
  • 46,490
  • Yes, it does: $\lim_{n\to\infty} \sqrt[n]{n^3} = \lim_{n\to\infty} (\sqrt[n]{n})^3 = \left(\lim_{n\to\infty} \sqrt[n]{n}\right)^3 = 1^3 = 1$, using the continuity of $x \mapsto x^3$, and the limit $\lim_{n\to\infty} \sqrt[n]{n} = 1$. – mechanodroid Aug 29 '17 at 19:31