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 $$\frac{1}{15}<0.0707<\frac{1}{2\cdot\sqrt{50}}<P_{50}<\frac{\sqrt{3}}{2\cdot \sqrt{101}}<0.0862<\frac{1}{10}\,.$$
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:
$2a>b$ ⇒ $2a^2>ab=\frac{1}{101}$⇒$a^2>\frac{1}{202}$⇒$a>\frac{1}{\sqrt{202}}>\frac{1}{\sqrt{225}}=\frac{1}{15}$
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 $n$.
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.
This is tedious and unsophisticated, but don't knock it, it works! :)
As well as answering the present question, it also answers the linked question, prove $\frac{1}{13}<\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{99}{100}<\frac{1}{12}$.
We need a preliminary lemma:
If $$ k = \left(1 - \frac{1}{52^2}\right)\left(1 - \frac{1}{56^2}\right)\cdots\left(1 - \frac{1}{96^2}\right), $$ then $$ \frac{383}{384} < k < \frac{1300}{1303}. $$ Rounding up and down, as appropriate, this is approximately $$ 0.997395 < k < 0.997698, $$ but of course we avoid using such calculations.
Proof. By the Weierstrass product inequality, we have $$ 1 - s < k < \frac{1}{1 + s}, $$ where $$ s = \frac{1}{52^2} + \frac{1}{56^2} + \cdots + \frac{1}{96^2} = \frac{1}{16}\left(\frac{1}{13^2} + \frac{1}{14^2} + \cdots + \frac{1}{24^2}\right). $$ Telescoping, \begin{align*} 16s & < \frac{1}{12\cdot13} + \frac{1}{13\cdot14} + \cdots + \frac{1}{23\cdot24} = \frac{1}{24}, \\ 16s & > \frac{1}{13\cdot14} + \frac{1}{14\cdot15} + \cdots + \frac{1}{24\cdot25} = \frac{12}{325}, \end{align*} therefore $$ 1 - \frac{1}{384} < k < \frac{1}{1 + \frac{3}{1300}}, $$ as required. $\square$
The number we wish to approximate is $$ P = \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{99}{100} = \frac{51\cdot53\cdot55\cdots97\cdot99}{2^{25}\cdot4\cdot8\cdot12\cdots96\cdot100} = kQ, $$ where \begin{gather*} Q = \frac{52^2\cdot56^2\cdots96^2\cdot99}{2^{75}\cdot25!} = \frac{13\cdot14\cdots24\cdot99}{2^{27}\cdot12!\cdot25} = \frac{13\cdot15\cdot17\cdot19\cdot21\cdot23\cdot99}{2^{21}\cdot6!\cdot25} \\ = \frac{7\cdot13\cdot17\cdot19\cdot23\cdot99}{2^{25}\cdot25}. \end{gather*} Therefore, using the bounds obtained for $k$ in the lemma, \begin{equation} \tag{$1$}\label{ineq:1} \frac{7\cdot13\cdot17\cdot19\cdot23\cdot33\cdot383}{2^{32}\cdot25} < P < \frac{7\cdot13^2\cdot17\cdot19\cdot23\cdot99}{2^{23}\cdot1303}. \end{equation} Although this is well within the range of easy hand calculation, I lazily used a calculator for most of the work (only doubling $2^{22}\cdot1303 = 5{,}465{,}178{,}112$ at the last stage by hand), to get \begin{equation} \tag{$2$}\label{ineq:2} \frac{8{,}544{,}456{,}921}{107{,}374{,}182{,}400} < P < \frac{870{,}062{,}193}{10{,}930{,}356{,}224}. \end{equation} Approximately, rounding up and down again, $$ 0.079576 < P < 0.079601. $$ One way to "simplify" \eqref{ineq:1} (although it seems simpler to me to calculate \eqref{ineq:2} and be done with it!) is as follows:
For the upper bound, observe that $7\cdot13\cdot19 = 1729 = 1 + 12^3$ (famously!), and $1302 = 2\cdot3\cdot7\cdot31$, so $$ \frac{7\cdot13\cdot19}{1303} < \frac{1728}{2\cdot3\cdot7\cdot31} = \frac{288}{217} < \frac{288}{216} = \frac{4}{3}, $$ (as one can now easily verify with hindsight), therefore $$ P < \frac{13\cdot17\cdot23\cdot99}{2^{21}\cdot3} < \frac{17\cdot300\cdot100}{2^{11}\cdot3\cdot1000} = \frac{85}{1024} < \frac{1}{12}. $$
For the lower bound, first simplify $Q$ by noticing that $7\cdot17\cdot19\cdot23 = 52003 > 52000 = 2^5\cdot5^3\cdot13$, whence $$ Q > \frac{5\cdot13^2\cdot99}{2^{20}} = \frac{5\cdot13\cdot1287}{2^{20}} > \frac{5\cdot13\cdot1280}{2^{20}} = \frac{5^2\cdot13}{2^{12}}, $$ and then, instead of using the lower bound from the lemma directly, weaken it to $P > \frac{64}{65}Q$, which gives $$ P > \frac{5}{64} > \frac{1}{13}. $$
For a better idea of the precision of this calculation, we can rewrite \eqref{ineq:2} as $$ \frac{489{,}609{,}908}{870{,}062{,}193} < \frac{1}{P} - 12 < \frac{4{,}840{,}699{,}348}{8{,}544{,}456{,}921}, $$ whence (this could be done by hand, although again I used a calculator) $$ \frac{1}{12.57} < P < \frac{1}{12.56}. $$
