Limit Value

kunny

Find the following limit value.

$\lim_{n\to\infty} \left(1-\frac{1}{1+\sqrt{2}}\right)\left(1-\frac{1}{\sqrt{2}+\sqrt{3}}\right)\cdots\cdots\left(1-\frac{1}{\...$

**Posted:** Thu Jan 27, 2005 9:17 pm

vuhung

Consider $a_n = \ln\left(1 - \frac{1}{\sqrt{n}+\sqrt{n+1}}\right)$ . It is easy to show that $a_n = -\frac{1}{2\sqrt{n}}+ O(1/\sqrt{n})$ and therefore $\sum a_n = -\infty$ . So the limit in the problem is 0.

**Posted:** Fri Jan 28, 2005 1:21 am

Moubinool

$c(n) =\left(1-\frac{1}{1+\sqrt{2}}\right)\left(1-\frac{1}{\sqrt{2}+\sqrt{3}}\right)\cdots\cdots\left(1-\frac{1}{\sqrt{n-1}+\s...$

Give an asymptotic expansion of c(n)

**Posted:** Fri Jan 28, 2005 11:46 am

kunny

In using $e^{-x}>1-x$ , for $k\geqq2, 0<1-\frac{1}{\sqrt{k-1}+\sqrt{k}}=1-(\sqrt{k-1}+\sqrt{k})<e^{\sqrt{k-1}-\sqrt{k}}$ ,yielding

$0<c(n)<e^{({1-\sqrt{2})+\scdots\cdots+(\sqrt{n-1}-\sqrt{n})=e^{1-\sqrt{n}}$

Therefore the desired limit value is $\lim_{n\to\infty} c(n)=0$

**Posted:** Fri Jan 28, 2005 8:49 pm