# Math Help - limit question

1. ## limit question

Here goes: I know that the limit as n goes to infinity of (
an) = A. I need to find the limit of the following as n goes to infinity:
(a1 + 2a2 + 3a3 + · · · + nan)
---------------------------------------
[n(n+1)] / 2

OK. I know the limit goes to A. And I know that (1 + 2 + 3 + ... + n) = [n(n+1)] / 2. So I want to say that this limit equals
(1 + 2 + 3 + ... + n)*(an)
------------------------, which would be just (an) and therefore its limit is A
[n(n+1)] / 2

Although I'm not sure if this works, since I would be multiplying each coefficient by an and not by a1, a2, or whatever.

Is there a different way to show this?

2. Apply Stolz-Cesaro theorem: Stolz-CesÃ*ro theorem - Wikipedia, the free encyclopedia

Let $x_n=1\cdot a_1+2\cdot a_2+\ldots+na_n, \ y_n=1+2+\ldots+n$.
$(y_n)$ is a sequence with positive terms, strictly ascending and unbounded.

Then $\displaystyle\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=\lim_{n\to\infty}\frac{(n+1)a_{n+1}}{n+1}=\li m_{n\to\infty}a_{n+1}=A$.

So, $\displaystyle\lim_{n\to\infty}\frac{x_n}{y_n}=A$

3. Originally Posted by BrainMan
Here goes: I know that the limit as n goes to infinity of (
an) = A. I need to find the limit of the following as n goes to infinity:
(a1 + 2a2 + 3a3 + · · · + nan)
---------------------------------------
[n(n+1)] / 2

OK. I know the limit goes to A. And I know that (1 + 2 + 3 + ... + n) = [n(n+1)] / 2. So I want to say that this limit equals
(1 + 2 + 3 + ... + n)*(an)
------------------------, which would be just (an) and therefore its limit is A
[n(n+1)] / 2

Although I'm not sure if this works, since I would be multiplying each coefficient by an and not by a1, a2, or whatever.

Is there a different way to show this?
Well, $a_1 + 2a_2 + ~ ... ~ + na_n \neq (1 + 2 + ~ ... ~ + n)a_n$
so this can't work.

This argument may not quite work either. Someone will have to double check me on this.

As you suggested, write this as
$\lim_{n \to \infty} \frac{a_1 + 2a_2 + ~ ... ~ + na_n}{1 + 2 + ~ ... ~ + n} = \lim_{n \to \infty} \frac{\sum_{k = 1}^nka_k}{\sum_{k = 1}^n k}$

Now, because $(a_n) = A$ we know that for some n = N (where N is sufficiently large) that $a_n$ is arbitrarily close to A. Similarly for a sufficiently large M (not necessarily the same as N, but we may take it to be as least as large as N) the fraction
$\frac{1}{1 + 2 + ~ ... ~}$ is arbitrarily close to $\frac{1}{M + (M + 1) + ~ ... ~}$. Thus:
$\lim_{n \to \infty} \frac{\sum_{k = 1}^nka_k}{\sum_{k = 1}^n k} = \lim_{n \to \infty} \frac{\sum_{k = N}^n kA}{\sum_{k = M}^n k}$

Now pick the larger of N and M and call it P. Then
$\lim_{n \to \infty} \frac{\sum_{k = 1}^nka_k}{\sum_{k = 1}^n k} = \lim_{n \to \infty} \frac{\sum_{k = P}^n kA}{\sum_{k = P}^n k} = A$

I think this will work, but I may have skipped crucial details.

-Dan