# Thread: help with a limit

1. ## help with a limit

I'm pretty sure this limit is zero, but I was hoping someone could help me show it with algebra:

Suppose that lim as n goes to infinity of (an) = 0. Let (bn) be a non negative sequence with lim as n goes to infinity of (b1 + b2 + · · · + bn1 + bn) = infinity.

What is the limit as n goes to infinity of:
(a1b1 + a2b2 + · · · + an1bn1 + anbn)
------------------------------------- ?
(b1 + b2 + · · · + bn1 + bn)

Should you divide everything by anbn or just bn? I'm not sure.

Thank you for any help.

2. We'll apply the Stolz-Cesaro lemma:
If $\displaystyle (x_n)$ and $\displaystyle (y_n)$ are two sequences such that $\displaystyle (y_n)$ is ascending and unbounded and if exists $\displaystyle \displaystyle\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=l$, then $\displaystyle \displaystyle\lim_{n\to\infty}\frac{x_n}{y_n}=l$.

Now, let $\displaystyle x_n=a_1b_1+a_2b_2+\ldots+a_nb_n, \ y_n=b_1+b_2+\ldots+b_n$

We have $\displaystyle y_n<y_{n+1}$ and $\displaystyle \lim_{n\to\infty}y_n=\infty$, so $\displaystyle (y_n)$ is unbounded.

$\displaystyle \displaystyle\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=\lim_{n\to\infty}\frac{a_{n+1}b_{n+1}}{b_{n+1 }}=\lim_{n\to\infty}a_{n+1}=0$.
So, $\displaystyle \lim_{n\to\infty}\frac{x_n}{y_n}=0$

3. I really appreciate the help. Is there a more complicated (or less direct) way to do it because I can't use that theorem in my class. It makes sense, but I looked at the proof of it and it's kind of out of my league.

Would you be able to prove the Stolz-Cesaro theorem with just elementary real analysis?

4. Don't worry about it. Thanks for the help.

5. Originally Posted by red_dog
We'll apply the Stolz-Cesaro lemma:
Can you teach me this lemma by posting it in this here.

I was going to add more tricks but I keep forgetting.