1. ## disprove convergence

Hi,

I have to show that, given a sequence (x_n) of real numbers with the property $|x_n|\geq n$ for infinitely many n, the sequence

$a_n=\frac{1}{n}\sum_{i=1}^n x_n$

does not converge. I tried a lot, but was not very successful.

Can I somehow show that $a_n$ can't be a Cauchy sequence?

Any ideas would be appreciated.

Thank you,
mili03

2. ## Re: disprove convergence

Assume that $a_n$ converge and try bounding $|a_{n+1}-a_n|$ from below using the fact that $\frac{1}{n}-\frac{1}{n(n+1)}=\frac{1}{n+1}$.

3. ## Re: disprove convergence

Originally Posted by emakarov
Assume that $a_n$ converge and try bounding $|a_{n+1}-a_n|$ from below using the fact that $\frac{1}{n}-\frac{1}{n(n+1)}=\frac{1}{n+1}$.
thx for the hint. Assuming $a_n$ converges: For each $\varepsilon>0$ there exists $n_0\in N$ so that for $n\geq n_0$
$|a_{n+1}-a_n|<\varepsilon$.

So if I find a lower bound, I am finished.

Now I should probably take a certain $n\geq n_0$ with $x_{n+1}\geq n+1$.

I have $|a_{n+1}-a_n|=|\frac{1}{n+1}\sum_{i=1}^{n+1}x_i-\frac{1}{n}\sum_{i=1}^n x_i|=|\frac{1}{n(n+1)}\sum_{i=1}^n x_i-\frac{1}{n+1}x_{n+1}|$

But I honestly speaking don't see how to be continued.

LG

4. ## Re: disprove convergence

If $\lim_{n\to\infty}a_n=A$, then there exists an $n_0$ such that $a_n\le 3A/2$ for all $n\ge n_0$.

$\left|\frac{x_{n+1}}{n+1}-\frac{1}{n+1}\frac{1}{n}\sum_{i=1}^n x_i\right|\ge\frac{|x_{n+1}|}{n+1}-\frac{|a_n|}{n+1}\ge1-\frac{3A}{2(n+1)}\xrightarrow[n\to\infty]{}\ 1$

Needs to be double-checked.

5. ## Re: disprove convergence

thank you, that's it! :-)

Originally Posted by emakarov
If $\lim_{n\to\infty}a_n=A$, then there exists an $n_0$ such that $a_n\le 3A/2$ for all $n\ge n_0$.
Here we require $|a_n|\le\frac{3|A|}{2}$.