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Let $x_1,x_2,x_3$,... be a sequence of nonnegative real numbers. Prove that

$\limsup_{n\to\infty}\frac{\sqrt{2({{x_1}^2+{x_2}^2+...+{x_n}^2})}}{n}\leq\limsup_{n\to\infty}\frac{x_n}{\sqrt{n}}$

I try: Let $M=\limsup_{n\to\infty}\frac{x_n}{\sqrt{n}}$. If $M=+\infty$,the inequality is clear. If $M<+\infty$, let $r\in \mathbb R$ and $r>M$. By $M_k$ theorem, $\exists k$ such that $M\leq M_k <r$ For $n>k$, we have $\frac{x_n}{\sqrt{n}}<r$, $\frac{x_{n-1}}{\sqrt{n}}<r$, . . . $\frac{x_{k+1}}{\sqrt{n}}<r$ , $\frac{x_k}{\sqrt{n}}<r$. Adding these, we get $\frac{{x_k}^2+...+{x_n}^2}{n}<(n-k)r^2$. So,$\frac{{x_1}^2+...+{x_n}^2}{n^2}<\frac{(n-k)r^2}{n}+\frac{{x_1}^2+...+{x_{k-1}}^2}{n^2}$. $\limsup_{n\to\infty}\frac{\sqrt{{x_1}^2+{x_2}^2+...+{x_n}^2}}{n}\leq \limsup_{n\to\infty}\sqrt{\frac{(n-k)r^2}{n}+\frac{{x_1}^2+...+{x_{k-1}}^2}{n^2}}=r$ Let $r\rightarrow M^+$, we get the inequality.

However,only $\limsup_{n\to\infty}\frac{\sqrt{{x_1}^2+{x_2}^2+...+{x_n}^2}}{n}\leq\limsup_{n\to\infty}\frac{x_n}{\sqrt{n}}$ is proved

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1 Answer 1

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Let $M = \limsup_n { x_n \over \sqrt{n}}$. Suppose $M<\infty$ otherwise the inequality is vacuous.

Then for all $\epsilon>0$, there exists some $N$ such that $x_n \le (M+\epsilon) \sqrt{n}$ for $n \ge N$.

Pick some $\epsilon>0$.

Then \begin{eqnarray} \sqrt{ { x_1^2+\cdots + x_n^2 \over n^2 } } &\le& \sqrt{ { x_1^2+\cdots + x_{N-1}^2 \over n^2 } +{ x_N^2+\cdots + x_n^2 \over n^2 } } \\ &\le & \sqrt{ { x_1^2+\cdots + x_{N-1}^2 \over n^2 } +(M+ \epsilon)^2{ N+\cdots + n \over n^2} } \\ &= & \sqrt{ { x_1^2+\cdots + x_{N-1}^2 \over n^2 } +(M+ \epsilon)^2{ {1 \over 2} (n-N+1) (N+n) \over n^2} } \\ \end{eqnarray} Taking the $\limsup$ of both sides gives $\limsup_n \sqrt{ { x_1^2+\cdots + x_n^2 \over n^2 } } \le {1 \over \sqrt{2}} (M+ \epsilon)$. Since $\epsilon>0$ was arbitrayy, we have $\limsup_n \sqrt{ { x_1^2+\cdots + x_n^2 \over n^2 } } \le { M \over \sqrt{2}}$.

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