$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot \cdot\frac{99}{100})<\frac{1}{10}$. 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 $P_n:=\prod\limits_{k=1}^n\,\dfrac{2k-1}{2k}$ for each integer $n\geq 1$.  We shall prove that $$\frac{2}{3\sqrt{2n}}<P_n<\frac{1}{\sqrt{2n}}\tag{*}$$
for every positive integer $n$, as suggested by Kemono Chen (I think there should be a factor $\dfrac{1}{\sqrt{2}}$ there so that (*) implies the OP's inequality for $n=50$).  The asymptotic behavior of $P_n$ is given here.
Observe that
$$P_n^2\leq \frac{1}{2^2}\,\left(\prod_{k=2}^n\,\frac{2k-1}{2k}\right)\,\left(\prod_{k=2}^n\,\frac{2k}{2k+1}\right)=\frac{1}{4}\,\prod_{k=3}^{2n}\,\frac{k}{k+1}=\frac{3}{4(2n+1)}.$$
In addition,
$$P_n^2\geq \frac{1}{2^2}\,\left(\prod_{k=2}^n\,\frac{2k-1}{2k}\right)\,\left(\prod_{k=2}^n\,\frac{2k-2}{2k-1}\right)=\frac{1}{4}\,\prod_{k=2}^{2n-1}\,\frac{k}{k+1}=\frac{2}{4(2n)}\,.$$
This shows that
$$\frac{1}{2\,\sqrt{n}}\leq P_n\leq \frac{\sqrt{3}}{2\,\sqrt{2n+1}}\,.$$
Both the inequality on the right and the inequality on the left have a unique equality case: $n=1$.  Note that this implies (*).  In particular, for $n=50$, we have
$$\frac{1}{15}<0.0707<\frac{1}{2\cdot\sqrt{50}}<P_{50}<\frac{\sqrt{3}}{2\cdot \sqrt{101}}<0.0862<\frac{1}{10}\,.$$
