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.
Compute $\lim_{n\to\infty} (3n)^{1/3} x_n$
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1Your 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 $\infty$. – 2012-09-30
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0Further, 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. – 2012-09-30
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0@ Harald Hanche-Olsen: thanks for your help. – 2012-09-30
1 Answers
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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0beautiful. Thanks! (+1). By the way, what is $a_{n}$? $x_n$, right? – 2012-09-30
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0I am sorry, you are right I corrected it. Glad I could help:) – 2012-09-30
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1I like that your approach of things is simple and useful. – 2012-09-30
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0I 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 $x_{n+1}$. – 2012-10-01