Ytukyg

Show that
$$frac{1}{15}<(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})<frac{1}{10}$$

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 $a=frac{1}{2}.frac{3}{4}.frac{5}{6}. . . .frac{99}{100}$

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:

$2a<frac{3}{2}b$⇒$2a^2<frac{3}{2}ab=frac{3}{2}.frac{3}{101}$⇒$a^2 >frac{4}{9times 101}$ ⇒ $a>frac{1}{15}$

Let $P_n:=prodlimits_{k=1}^n,dfrac{2k-1}{2k}$ for each integer $ngeq 1$. We shall prove that $$frac{2}{3sqrt{2n}}<P_n<frac{1}{sqrt{2n}}tag{*}$$
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
$$P_n^2leq frac{1}{2^2},left(prod_{k=2}^n,frac{2k-1}{2k}right),left(prod_{k=2}^n,frac{2k}{2k+1}right)=frac{1}{4},prod_{k=3}^{2n},frac{k}{k+1}=frac{3}{4(2n+1)}.$$
In addition,
$$P_n^2geq frac{1}{2^2},left(prod_{k=2}^n,frac{2k-1}{2k}right),left(prod_{k=2}^n,frac{2k-2}{2k-1}right)=frac{1}{4},prod_{k=2}^{2n-1},frac{k}{k+1}=frac{2}{4(2n)},.$$
This shows that
$$frac{1}{2,sqrt{n}}leq P_nleq frac{sqrt{3}}{2,sqrt{2n+1}},.$$
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
$$frac{1}{15}<0.0707<frac{1}{2cdotsqrt{50}}<P_{50}<frac{sqrt{3}}{2cdot sqrt{101}}<0.0862<frac{1}{10},.$$

Let $S_n=prod_{k=1}^nfrac{2k-1}{2k}$. From Wallis' product, we have
$$prod_{k=1}^inftyleft(frac{2k}{2k-1}cdotfrac{2k}{2k+1}right)=frac{pi}{2}.$$
Since each term in the product above is greater than $1$, this shows that
$$prod_{k=1}^nleft(frac{2k}{2k-1}cdotfrac{2k}{2k+1}right)<frac{pi}{2}$$
for all $n$. That is,
$$(2n+1)S_n^2=(2n+1)prod_{k=1}^nleft(frac{2k-1}{2k}right)^2=prod_{k=1}^nleft(frac{2k-1}{2k}cdotfrac{2k+1}{2k}right)>frac{2}{pi}.$$
Therefore,
$$S_n>frac{1}{sqrt{pileft(n+frac12right)}}.$$

Similarly, Wallis' product also implies that
$$prod_{k=2}^inftyleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)=frac{4}{pi}.$$
Since each term in the product above is greater than $1$, this shows that
$$prod_{k=2}^nleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)<frac{4}{pi}$$
for all $n$. That is,
$$2nS_n^2=2nprod_{k=1}^nleft(frac{2k-1}{2k}right)^2=frac12prod_{k=2}^nleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)<frac12left(frac{4}{pi}right).$$
Therefore,
$$S_n<frac{1}{sqrt{pi n}}.$$ Hence,
$$frac{1}{sqrt{pileft(n+frac12right)}}<S_n<frac1{sqrt{pi n}}$$
for every $n$.

Just a thought, that may be worth mentioning:

The expression:

$$p=(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})$$

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 $frac{1}{13}<frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdotsfrac{99}{100}<frac{1}{12}$.

We need a preliminary lemma:

If
$$
k = left(1 - frac{1}{52^2}right)left(1 - frac{1}{56^2}right)cdotsleft(1 - frac{1}{96^2}right),
$$
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,
$$
frac{12}{325} =
frac{1}{13cdot14} + frac{1}{14cdot15} + cdots + frac{1}{24cdot25}
< 16s <
frac{1}{12cdot13} + frac{1}{13cdot14} + cdots + frac{1}{23cdot24}
= frac{1}{24},
$$
therefore
$$
1 - frac{1}{384} < k < frac{1}{1 + frac{3}{1300}},
$$
as required. $square$

The number we wish to approximate is
$$
P = frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdotsfrac{99}{100}
= frac{51cdot53cdot55cdots97cdot99}{2^{25}cdot4cdot8cdot12cdots96cdot100}
= kQ,
$$
where
begin{gather*}
Q = frac{52^2cdot56^2cdots96^2cdot99}{2^{75}cdot25!}
= frac{13cdot14cdots24cdot99}{2^{27}cdot12!cdot25}
= frac{13cdot15cdot17cdot19cdot21cdot23cdot99}{2^{21}cdot6!cdot25} \
= frac{7cdot13cdot17cdot19cdot23cdot99}{2^{25}cdot25}.
end{gather*}
Therefore, using the bounds obtained for $k$ in the lemma,
begin{equation}
tag{$1$}label{ineq:1}
frac{7cdot13cdot17cdot19cdot23cdot33cdot383}{2^{32}cdot25}
< P <
frac{7cdot13^2cdot17cdot19cdot23cdot99}{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}
or approximately
$$
0.079576 < P < 0.079601.
$$
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 $7cdot13cdot19 = 1729 = 1 + 12^3$ (famously!), and $1302 = 2cdot3cdot7cdot31$, so
$$
frac{7cdot13cdot19}{1303} < frac{1728}{2cdot3cdot7cdot31} = frac{288}{217} < frac{288}{216} = frac{4}{3},
$$
(as one can now easily verify with hindsight), therefore
$$
P < frac{13cdot17cdot23cdot99}{2^{21}cdot3} < frac{17cdot300cdot100}{2^{11}cdot3cdot1000} = frac{85}{1024} < frac{1}{12}.
$$

For the lower bound, first simplify $Q$ by noticing that $7cdot17cdot19cdot23 = 52003 > 52000 = 2^5cdot5^3cdot13$, whence
$$
Q > frac{5cdot13^2cdot99}{2^{20}} = frac{5cdot13cdot1287}{2^{20}} > frac{5cdot13cdot1280}{2^{20}} = frac{5^2cdot13}{2^{12}},
$$
and then, instead of using the lower bound from the lemma directly, weaken it to $P > frac{64}{65}Q$, which gives
$$
P > frac{5}{64} > frac{1}{13}.
$$