1. ## lim[a_n]=a, then lim[(a_1+a_2+...+a_n)/n]=a

How can I prove that if $\displaystyle a_{n}\to a$, then $\displaystyle b_{n}=\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}\to a$ ?

Can I apply the definition for Cauchy sequences here, or is this a proof for the conditions of Cauchy sequences?

Thanks

2. Originally Posted by tunaaa
How can I prove that if $\displaystyle a_{n}\to a$, then $\displaystyle b_{n}=\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}\to a$ ?

Can I apply the definition for Cauchy sequences here, or is this a proof for the conditions of Cauchy sequences?

Thanks

I supose you must have meant $\displaystyle b_{n}=\frac{1+\sqrt{a_2}+\sqrt[3]{a_3}+...+\sqrt[n]{a_n}}{n}\to a$ , and thus we must impose the condition that $\displaystyle \forall\,n\in\mathbb{N}\,,\,\,a_n\geq 0$...or something like this. Check it.

Tonio

3. Sorry, I was thinking of another question - I actually meant:

$\displaystyle b_{n}=\frac{a_1+a_2+...+a_n}{n}\to a$

Thanks

4. Hello,

Look for Cesàro's theorem (or Cesàro's mean) in google.

5. Originally Posted by tunaaa
Sorry, I was thinking of another question - I actually meant:

$\displaystyle b_{n}=\frac{a_1+a_2+...+a_n}{n}\to a$

Thanks

This is what you need: http://www.math.ksu.edu/~nagy/snippets/stolz-cesaro.pdf

The exact result you're trying to prove is in page 2.

Tonio

6. Originally Posted by tonio
This is what you need: http://www.math.ksu.edu/~nagy/snippets/stolz-cesaro.pdf

The exact result you're trying to prove is in page 2.

Tonio
Thanks, but is there another proof that doesn't involve lim sup or lim inf?

7. Originally Posted by tunaaa
Thanks, but is there another proof that doesn't involve lim sup or lim inf?

What do you care about that?? They all are the same if the sequence converges, which is YOUR case...
You can even modify the proof there by assuming your seq. converges and all the time lim sup and lim inf are the same.

Tonio

8. Originally Posted by tonio
What do you care about that?? They all are the same if the sequence converges, which is YOUR case...
You can even modify the proof there by assuming your seq. converges and all the time lim sup and lim inf are the same.

Tonio
Sorry, it's just that we've not covered lim sup and lim inf yet.

9. Originally Posted by tunaaa
Sorry, it's just that we've not covered lim sup and lim inf yet.
Let $\displaystyle \epsilon>0$. Then $\displaystyle \exists N$ such that $\displaystyle \forall n>N$, $\displaystyle |a_n-a|<\epsilon$, so let $\displaystyle K=\sum_{n=1}^{N-1}a_n$, and we have

$\displaystyle \lim_{n\to\infty}\frac{K+(n-N)(a-\epsilon)}{n}\leq\lim_{n\to\infty}\frac{a_1+a_2+.. .+a_n}{n}\leq\lim_{n\to\infty}\frac{K+(n-N)(a+\epsilon)}{n}$

10. Originally Posted by redsoxfan325
we have

$\displaystyle \lim_{n\to\infty}\frac{K+(n-N)(a-\epsilon)}{n}\leq\lim_{n\to\infty}\frac{a_1+a_2+.. .+a_n}{n}\leq\lim_{n\to\infty}\frac{K+(n-N)(a+\epsilon)}{n}$
Thanks, but can you explain how you arrived at this stage please?

11. Originally Posted by tunaaa
Thanks, but can you explain how you arrived at this stage please?
$\displaystyle \frac{(a_1+a_2+...+a_N)+(a_{N+1}+...+a_n)}{n}$

We know that for $\displaystyle n>N$, $\displaystyle |a_n-a|<\epsilon\implies a-\epsilon<a_n<a+\epsilon$.

So I called the sum of the first $\displaystyle N$ terms $\displaystyle K$ (because the sum of a finite number of terms is just a number), and since there are $\displaystyle n-N$ terms left, each of which is bounded below by $\displaystyle a-\epsilon$ (and above by $\displaystyle a+\epsilon$), the inequality follows.