Question in my test today

1. Dec 9, 2004

quasar987

There was this question in my analysis exam today. I have a feeling it should be easy but no one I've asked knew how to do it.

We associate to the sequence $\{a_n\}$ the sequence defined by

$$b_n=\frac{a_1+a_2+...+a_n}{n}$$

Show that if $\{a_n\}$ converges towards a, then $\{b_n\}$ converges towards a.

I realised that

$$b_n=\frac{\sum_{n=1}^{\infty} a_n}{n}$$

or even

$$b_n=\frac{\sum_{k=1}^{n} a_k}{\sum_{k=1}^{n} 1}$$

but all my attemps involving epsilon-delta, convergence tests, Cauchy convergence "caracterisation", etc. failed. Please tell me how to do this. Thanks a lot.

2. Dec 9, 2004

Hurkyl

Staff Emeritus
It often helps to split convergent sequences into two parts, an initial part, and the rest, because you can decide how much "the rest" resembles the limit.

3. Dec 9, 2004

fourier jr

let $$\epsilon > 0$$. then since a_n converges there exists N>0 such that $$a - \epsilon < a_n < a + \epsilon$$.

now $$b_n = \frac{a_1 + a_2 + ... + a_n}{n}$$ for n$$\geq$$

so $$b_n = \frac{a_1 + a_2 + ... + a_N}{n} + \frac{a_1 + a_2 + ... + a_n}{n}$$

& since $$\frac{(n-N)(a-\epsilon)}{n} < \frac{a_N+1 + ... + a_n}{n} < \frac{(n-N)(a+\epsilon)}{n}$$,

$$\frac{a_1 + a_2 + ... + a_N}{n} + \frac{(n-N)(a-\epsilon)}{n} < b_n < \frac{a_1 + a_2 + ... + a_N}{n} + \frac{(n-N)(a+\epsilon)}{n}$$

now using that fact that $$\limsup a_n \leq \limsup b_n$$ if $$a_n \leq b_n$$ on the previous inequality we get

$$\limsup ( \frac{a_1 + a_2 + ... + a_N}{n} + \frac{(n-N)(a-\epsilon)}{n} ) \leq \limsup b_n \leq \limsup \frac{a_1 + a_2 + ... + a_N}{n} + \frac{(n-N)(a+\epsilon)}{n}$$.

since $$\lim ( \frac{a_1 + a_2 + ... + a_N}{n} + \frac{(n-N)(a-\epsilon)}{n} ) = 0 + (a - \epsilon$$ )
and $$\lim ( \frac{a_1 + a_2 + ... + a_N}{n} + \frac{(n-N)(a+\epsilon)}{n} ) = 0 + (a + \epsilon$$ )

we can conclude that
a - $$\epsilon \leq \limsup b_n \leq a+\epsilon$$, for any $$\epsilon > 0$$

since epsilon is arbitrary, $$a \leq \limsup b_n \leq a$$, so $$a = \limsup b_n$$, similarily for liminf, and the result follows from the limsup & liminf being equal

Last edited: Dec 9, 2004
4. Dec 9, 2004

quasar987

holy cow!

I now feel my feeling was injustified. It's a beautiful proof and very instructive; a gazilion thanx fourier!

5. Dec 10, 2004

fourier jr

there are a couple lines that need fixing but i wonder if it would be worth the trouble or if people can still follow it. i need to get used to the tex-ing

6. Dec 10, 2004

quasar987

It's good enough. And if I could understand it, the others must have too.