Matematické Fórum

↑ stuart clark:
Hi Stuart Clark,
There Is  $\underset{n\to \mathrm{\infty }}{lim}exp(\frac{\sum _{k=0}^{n}logC(n;k)}{n(n+1)})$
When I use Stolz Cesar Theorem for limit (), I get
$(n+2)(n+1)-(n+1)n=2(n+1)$
$\sum _{k=0}^{n+1}logC(n+1;k)-\sum _{k=0}^{n}logC(n;k)=-\sum _{k=0}^{n}log(1-\frac{k}{n+1})$
There Is then $(-\frac{1}{2})\cdot {\int }_0^{1}log(1-x)ex=\frac{1}{2}$
Then the result Is $\sqrt{e}$

Hi ↑ krakonoš:, ↑ stuart clark:,
just a comment: you can avoid Stolz-Cesaro by some easy algebra.

First of all, in this case, you can get rid of the binomial numbers and factorials by simply switching the order of taking the product, e.g.:
$\prod _{k=0}^{n}k!=\prod _{k=0}^n\prod _{j=1}^{k}j=\prod _{j=1}^n{j}^{n-j+1}$
This leads quickly to
$(\genfrac{}{}{0ex}{}{n}{0})\cdot (\genfrac{}{}{0ex}{}{n}{1})\cdot (\genfrac{}{}{0ex}{}{n}{2})\cdots \cdots (\genfrac{}{}{0ex}{}{n}{n})=\prod _{k=1}^n{k}^{2k-1-n}$
and you are left with the limit of
$\frac{1}{n(n+1)}\sum _{k=1}^n(2k-1-n)\ln k$
which, applying the Riemann integration technique as krakonos suggested, converges to
${\int }_0^1(2x-1)\ln x=\frac{1}{2}.$