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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0@ Harald Hanche-Olsen: thanks $f$or 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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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