$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot \cdot\frac{99}{100})<\frac{1}{10}$. 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. 
 A: 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
$$\frac{1}{15}<0.0707<\frac{1}{2\cdot\sqrt{50}}<P_{50}<\frac{\sqrt{3}}{2\cdot \sqrt{101}}<0.0862<\frac{1}{10}\,.$$
A: Let $a=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}. . . .\frac{99}{100}$
and:
$b=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}. . . \frac{100}{101}$
It is clear that $a<b$ because each factor of $a$ is less than its corresponding factor in $b$ :
$\frac{1}{2}<\frac{2}{3}, \frac{3}{4}<\frac{4}{5}. . . \frac{99}{100}<\frac{100}{101}$
⇒ $a^2 < ab=(\frac{1}{2}.\frac{2}{3}).(\frac{3}{4}.\frac{4}{5}). . . .(\frac{99}{100}.\frac{100}{101})=\frac{1}{101}$
$a^2<\frac{1}{101}$⇒$a<\frac{1}{\sqrt{101}}<\frac{1}{10}$
Also:
$2a=\frac{3}{4}.\frac{5}{6}. . . .\frac{99}{100}$
$\frac{3}{2}.b=\frac{4}{5}.\frac{6}{7}. . . \frac{100}{101}$
$2a<\frac{3}{2}b$⇒$2a^2<\frac{3}{2}ab=\frac{3}{2}.\frac{3}{101}$ 
Or $a^2<\frac{3}{4}ab=\frac{3}{4}.\frac{3}{101}$
Since $9>4$ then  $a^2 >\frac{4}{9\times 101}$ and therefore:
$a>\frac{1}{15}$
A more reliable reasoning is given as a comment for this part:
$2a>b$ ⇒ $2a^2>ab=\frac{1}{101}$⇒$a^2>\frac{1}{202}$⇒$a>\frac{1}{\sqrt{202}}>\frac{1}{\sqrt{225}}=\frac{1}{15}$
A: 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 $n$.


A: 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.
