$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot \cdot\frac{99}{100})<\frac{1}{10}$. 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. 

Just a thought, that may be worth mentioning:
The expression:
$$p=(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot\cdot\frac{99}{100})$$
We could use the following identities:
Product of $n$ odd numbers = 
$$p_o = \frac{(2n!)}{(n!)2^{n}}$$
Product of $n$ even numbers = 
$$p_e = (n!)2^{n}$$
The first $4$ terms of $p$ =
$$\frac{1.3.5.7}{2.4.6.8}=\frac{105}{1152}=0.2734$$
We may write $p$ as:
$$p=\frac{odd_numbers}{even_numbers}=\frac{p_o}{p_e}=$$
$$p=\frac{(2n)!}{((n!)2^{n})^2}$$
To prove $p<\frac{1}{10}$, we know that (2n)! < n! (for n>1), so we can write:
$$p<\frac{(n!)}{((n!)2^{n})^2}$$
$$p<\frac{1}{(n!)({n})^2}$$
We could conclude that, for $n >=2$
$$p<\frac{1}{10}$$
for the lower bound, any of the other solution presented may be considered or check stovf-math p between 1/13 and 1/15 may also be considered.
