Proving Convergence of Sequences Using the Mean Value Theorem

[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 realized 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.

 

[Hurkyl]

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.

 

[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

 

[quasar987]

holy cow!

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

 

[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

 

[quasar987]

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