Proving limit

**zadir** · April 3rd 2011, 09:57 AM

Let be $x_1 \geq x_2 \geq ... \geq x_n \geq ...$ positive numbers and $\sum _{n=1}^{\infty}x_n < \infty.$ Prove that $\lim_{n \to \infty}nx_n =0.$

Thank you very much in advance!

**girdav** · April 3rd 2011, 10:30 AM

We have $0\leq nx_{2n}\leq x_{n+1}+\ldots +x_{2n}=s_{2n}-s_n$ where $s_n:=\sum_{k=1}^n x_k$ . Hence the limit of the subsequence $\left\{2nx_{2n}\right\}$ is $0$ . Now show that the limit is $0$ for the subsequence $\left\{(2n+1)x_{2n+1}\right\}$ .

**chisigma** · April 3rd 2011, 12:10 PM

Originally Posted by zadir

Let be $x_1 \geq x_2 \geq ... \geq x_n \geq ...$ positive numbers and $\sum _{n=1}^{\infty}x_n < \infty.$ Prove that $\lim_{n \to \infty}nx_n =0.$

Thank you very much in advance!

The series with positive terms $\displaystyle \sum_{n=1}^{\infty} x_{n}$ converges and that means that there is an $\varepsilon >0$ and an $\alpha>0$ such that for n 'large enough' is...

$\displaystyle x_{n} < \frac{\alpha}{n^{1+\varepsilon}}$ (1)

From (1) it follows that...

$\displaystyle \lim_{n \rightarrow \infty} n\ x_{n} =0$ (2)

Kind regards

$\chi$ $\sigma$

**girdav** · April 3rd 2011, 01:03 PM

Chisigma : are you sure that your property is true ? What about $x_n :=\dfrac 1{n(\ln n)^2}$ ?

**zadir** · April 6th 2011, 12:38 AM

Thank you very much for all your answers!

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