$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot \cdot\frac{99}{100})<\frac{1}{10}$. Show that
$$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot\cdot\frac{99}{100})<\frac{1}{10}$$
My attempt: This  problem is from a text book where is introduced as: https://en.wikipedia.org/wiki/Generalized_mean 
This wouldn't be a problem if I knew the sum or a product of the given numbers. Then I will use AG inequality, but I don't know how. 

Let $S_n=\prod_{k=1}^n\frac{2k-1}{2k}$.  From Wallis' product, we have
$$\prod_{k=1}^\infty\left(\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\right)=\frac{\pi}{2}.$$
Since each term in the product above is greater than $1$, this shows that
$$\prod_{k=1}^n\left(\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\right)<\frac{\pi}{2}$$
for all $n$.  That is,
$$(2n+1)S_n^2=(2n+1)\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^2=\prod_{k=1}^n\left(\frac{2k-1}{2k}\cdot\frac{2k+1}{2k}\right)>\frac{2}{\pi}.$$
Therefore,
$$S_n>\frac{1}{\sqrt{\pi\left(n+\frac12\right)}}.$$
Similarly, Wallis' product also implies that
$$\prod_{k=2}^\infty\left(\frac{2k-1}{2k-2}\cdot\frac{2k-1}{2k}\right)=\frac{4}{\pi}.$$
Since each term in the product above is greater than $1$, this shows that
$$\prod_{k=2}^n\left(\frac{2k-1}{2k-2}\cdot\frac{2k-1}{2k}\right)<\frac{4}{\pi}$$
for all $n$.  That is,
$$2nS_n^2=2n\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^2=\frac12\prod_{k=2}^n\left(\frac{2k-1}{2k-2}\cdot\frac{2k-1}{2k}\right)<\frac12\left(\frac{4}{\pi}\right).$$
Therefore,
$$S_n<\frac{1}{\sqrt{\pi n}}.$$  Hence,
$$\frac{1}{\sqrt{\pi\left(n+\frac12\right)}}<S_n<\frac1{\sqrt{\pi n}}$$
for every $n$.


