# Compute $\lim_{n\to\infty} (3n)^{1/3} x_n$

Let $(x_{n})_{n \ge 1}$ be a sequence of real numbers with $$\lim_{n\to\infty} x_n \sum_{k=1}^{n}x^2_{k}=1$$ Compute $$\lim_{n\to\infty} (3n)^{1/3} x_n$$ My guess so far is that $x_{n}$ tends to $0$ and the sum tends to $\infty$. Could you help here? Thanks.

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Your guess is certainly good: It $x\not\to0$ then the sum diverges, and the combination of these two facts contradicts the stated limit. Now since $x\to0$, the sum must diverge to $\infty$. – Harald Hanche-Olsen Sep 30 '12 at 15:23
Further, before trying to solve the general problem, you might try to guess the answer, by trying to put $x_n=Cn^\gamma$ for some constants $C$ and $\gamma$. You need $\gamma\ge-1/2$ for the sum to diverge; then the sum is asymptotically $C^2\int_0^n x^{2\gamma}\,dx$, and you quickly end up with $\gamma=-1/3$. You can now solve for $C$ and compute the requested limit for this example. – Harald Hanche-Olsen Sep 30 '12 at 15:31
@ Harald Hanche-Olsen: thanks for your help. – user 1618033 Sep 30 '12 at 17:41

By Harald's argument we have that $$x_n \rightarrow 0, \, \, S_n=\sum_{k=1}^{n}x_k^2 \rightarrow \infty$$ We use Stolz's lemma to show $$\lim_n \frac{3n}{S_n^3}=1$$ And we would be done after that, indeed $$\lim_n 3n x_n^3=\lim_n \frac{3nx_n^3}{x_n^3S_n^3}=\lim_n \frac{3n}{S_n^3}=1$$ and so by continuity of $f(x)= x^\frac{1}{3}$ we are done.

Now let's prove our claim $$\lim_n \frac{3n}{S_n^3}=\lim_n \frac{3(n+1)-3n}{S_{n+1}^3- S_n^3}=$$ $$=\frac{3}{(S_{n+1}-S_{n})( S_{n+1}^2 + S_{n+1}S_n + S_n^2)}=$$ $$=\frac{3}{x_{n+1}^2( S_{n+1}^2 + S_{n+1}S_n + S_n^2)}=$$ $$=\frac{3}{x_{n+1}^2 S_{n+1} ^2( 1+\frac{S_n}{S_{n+1} } + \frac{S_n}{S_{n+1}})^2}=$$

Now $$\lim_n\frac{S_n}{S_{n+1}}=\lim_{n}\left (1 -\frac{x_{n+1}^2}{S_{n+1}} \right )=1$$ Because $$\frac{x_n ^2}{S_n}=\frac{x_n ^3}{S_n x_n}$$ So we are left

$$\lim_n \frac{3n}{S_n^3}=\lim_n\frac{3}{x_{n+1}^2 S_{n+1}^2 } \frac{1}{3}=1$$

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beautiful. Thanks! (+1). By the way, what is $a_{n}$? $x_n$, right? – user 1618033 Sep 30 '12 at 17:34
I am sorry, you are right I corrected it. Glad I could help:) – clark Sep 30 '12 at 17:38
I like that your approach of things is simple and useful. – user 1618033 Sep 30 '12 at 17:41
I think it is $\lim_n\frac{S_n}{S_{n+1}}=\lim_{n}\left (1 -\frac{x^2_{n+1}}{S_{n+1}} \right )=1$. I mean $x^2_{n+1}$ instead of $x_{n+1}$. – user 1618033 Oct 1 '12 at 7:35
Thank you, corrected:) – clark Oct 1 '12 at 7:40