1. ## two more limits

Could someone show me these?

Show that if lim as n goes to infinity (an) = A, then lim as n goes to infinity of
na1 + (n-1)a2 + ... + 2an-1 + an
----------------------------- = (A/2)
n^2

and

Show that for p in the natural numbers, the lim as n goes to infinity of
(1^p) + (2^p) + ... + (n^p)
------------------------- = 1/(p+1)
n^(p+1)

Thanks in advance for any suggestions or help.

2. All these limits you posted here are of the same kind and all can be solved by using the Stolz-Cesaro theorem. I don't know why your teacher gave you these problems if he didn't show you this theorem.

Now, for the first one.
Let $x_n=na_1+(n-1)a_2+\ldots+2a_{n-1}+a_n, \ y_n=n^2$.
$(y_n)$ is a positive sequence, strictly ascending and unbounded.

$\displaystyle\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=$
$\displaystyle=\lim_{n\to\infty}\frac{a_1+a_2+\ldot s+a_{n+1}}{2n+1}=$
$\displaystyle=\lim_{n\to\infty}\frac{a_1+a_2+\ldot s+a_{n+1}}{n+1}\cdot\frac{n+1}{2n+1}=A\cdot\frac{1 }{2}=\frac{A}{2}$.

Here i used the following result:
If $\lim_{n\to\infty}a_n=A$ then $\displaystyle\lim_{n\to\infty}\frac{a_1+a_2+\ldots +a_n}{n}=A$

Then $\displaystyle\lim_{n\to\infty}\frac{x_n}{y_n}=\fra c{A}{2}$.

The second one:
Let $x_n=1^p+2^p+\ldots+n^p, \ y_n=n^{p+1}$.
$(y_n)$ is a positive sequence, strictly ascending and unbounded.

$\displaystyle\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=$
$\displaystyle=\lim_{n\to\infty}\frac{(n+1)^p}{(n+1 )^{p+1}-n^{p+1}}=\lim_{n\to\infty}\frac{n^p+C_p^1n^{p-1}+C_p^2n^{p-2}+\ldots+1}{n^{p+1}+C_{p+1}^1n^p+C_{p+1}^2n^{p-1}+\ldots+1-n^{p+1}}=$
$\displaystyle=\lim_{n\to\infty}\frac{n^p+\ldots+1} {(p+1)n^p+\ldots+1}=\frac{1}{p+1}$.

So, $\displaystyle\lim_{n\to\infty}\frac{x_n}{y_n}=\fra c{1}{p+1}$.

3. Thanks a lot for helping.

Um, this one is quite similar, but I can't seem to simplify it to the correct limit. Bare with me as I type this out.
lim as n goes to infinity of (an) = A
Find the limit as n goes to infinity of
[(n)*a1 +
(1)

(n)*a2 + ........ +
(2)

(n)*an ]
(n)
--------------------------
2^(n).

where
(n) = n!/[(n!)(n-n)!] (i.e. a combination without repetition)
(n)
The limit equals A.
It seems I can't use the theorem because
(n)
(n+1)
does not exist.

I tried something else, but I'm getting tired of writing like this. Could you show me how to do this one as well?

I really appreciate you helping.

P.S. I'm sorry about the typing. I need to learn latex.

4. $\displaystyle\lim_{n\to\infty}\frac{C_n^1a_1+C_n^2 a_2+\ldots+C_n^na_n}{2^n}=$
$\displaystyle=\lim_{n\to\infty}\frac{C_{n+1}^1a_1+ C_{n+2}^2a_2+\ldots+C_{n+1}^na_n+C_{n+1}^{n+1}a_{n +1}-C_n^1a_1-C_n^2a_2-\ldots-C_n^na_n}{2^{n+1}-2^n}=$
$\displaystyle=\lim_{n\to\infty}\frac{a_1(C_{n+1}^1-C_n^1)+a_2(C_{n+1}^2-C_n^2)+\ldots+a_n(C_{n+1}^n-C_n^n)+a_{n+1}}{2^n}=$
$\displaystyle=\lim_{n\to\infty}\frac{C_n^0a_1+C_n^ 1a_2+\ldots+C_n^{n-1}a_n+C_n^na_{n+1}}{C_n^0+C_n^1+\ldots+C_n^n}=A$

In the last limit, the fraction represents the arithmetic mean of the numbers
$a_1,\underbrace{a_2, a_2,\ldots,a_2}_{C_n^1},\underbrace{a_3,a_3,\ldots ,a_3}_{C_n^2},\ldots,\underbrace{a_{n+1},a_{n+1},\ ldots,a_{n+1}}_{C_n^n}$