1/2 * 3/4 * 5/6 * ... * 99/100 less than 1/10
<meta name="title" content="An inequality: 1/2*3/4*5/6* ... 99/100
An Inequality:
$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{99}{100} < \frac{1}{10}$
A product of fractions $\displaystyle \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{2n-1}{2n}$ is on the left-hand side of several inequalities: one with a beautiful proof, one that strengthens the former but is virtually impossible to prove, and a third, even stronger, with an elementary proof.
Try your hand with the simplest variation:
$\displaystyle \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{99}{100} \lt\frac{1}{10}.$
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Denote the left-hand side of the inequality A:
$\displaystyle A = \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots \cdot\frac{99}{100}.$
And introduce its nemesis $B$:
$\displaystyle B = \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\ldots\cdot\frac{98}{99}.$
Factor by factor, the fractions in $B$ exceed those in $A:$
$\displaystyle \frac{2}{3} \gt \frac{1}{2},$ $\displaystyle \frac{4}{5} \gt \frac{3}{4},\ldots,\frac{98}{99} \gt \frac{97}{98},$ $\displaystyle 1 \gt \frac{99}{100}.$
From this it follows that $A \lt B.$ Note that, due to the choice of $B,$ in the product $AB$ most of the terms cancel out: $\displaystyle AB = \frac{1}{100}.$ From here,
$\displaystyle A^{2} \lt
AB = \frac{1}{100},$
which, with one additional step, proves (1).
This proof suggests that (1) is in fact just a special case of a more general inequality
$\displaystyle \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots \cdot\frac{2n-1}{2n} \lt \frac{1}{\sqrt{2n}},$
whose proof is a slight modification of the above with $A$ and $B$ defined as
$\displaystyle A(n) = \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots
\cdot\frac{2n-1}{2n},\\
\displaystyle B(n) = \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\ldots \cdot\frac{2n-2}{2n-1}.$
As we shall see shortly, (1) and (2) are quite weak: $A(n)$ has a much better bound, viz.
$\displaystyle A(n) \lt\frac{1}{\sqrt{3n+1}}.$
(3) supplies an edifying curiosity. By itself, it is easily proved by . However, its weakened version
$\displaystyle A(n) \lt\frac{1}{\sqrt{3n}},$
as far as I know, does not submit to an inductive proof. Try it, by all means. (3) and (3') are often quoted as a pair of problems of which the harder one has a simpler proof.
Meanwhile here's a .
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To remind,
$\displaystyle A(n) = \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots
\cdot\frac{2n-1}{2n}$
and we wish to prove (3): $\displaystyle A(n) \lt\frac{1}{\sqrt{3n+1}}.$ For $n = 1,$ we have
$\displaystyle A(1) = \frac{1}{2} = \frac{1}{\sqrt{3\cdot 1+1}}.$
But already for $n = 2,$
$\displaystyle A(2) = \frac{1}{2}\cdot\frac{3}{4} = \frac{3}{8} \lt\frac{1}{\sqrt{7}} = \frac{1}{\sqrt{3\cdot 2+1}},$
because upon squaring $\displaystyle \frac{9}{64} \lt\frac{1}{7},$ for $7\cdot 9 = 63 \lt 64.$ Thus let's proceed with the inductive step and assume that (3) holds for $n = k:$
$\displaystyle A(k) \lt\frac{1}{\sqrt{3k+1}}.$
We are going to prove that, for $n = k+1,$ (3) also holds
$\displaystyle A(k+1) \lt\frac{1}{\sqrt{3(k+1)+1}} =\frac{1}{\sqrt{3k+4}}.$
Since $\displaystyle A(k+1) = A(k)\cdot\frac{2k+1}{2k+2},$ (4) implies
$\displaystyle A(k+1) \lt\frac{2k+1}{2k+2}\cdot\frac{1}{\sqrt{3k+1}}.$
Now square the right hand side in (6):
$\displaystyle \begin{align}
\left(\frac{2k+1}{2k+2}\cdot\frac{1}{\sqrt{3k+1}}\right)^{2}&= \frac{(2k+1)^{2}}{(2k+2)^{2}(3k+1)}\\
&= \frac{(2k+1)^{2}}{12k^{3} + 28k^{2} + 20k + 4}\\
&= \frac{(2k+1)^{2}}{(12k^{3} + 28k^{2} + 19k + 4) + k}\\
&= \frac{(2k+1)^{2}}{(2k+1)^{2}(3k+4) + k}\\
&\lt\frac{(2k+1)^{2}}{(2k+1)^{2}(3k+4)}\\
&= \frac{1}{3k+4},
\end{align}$
which is exactly the right-hand side of (5) and proves (6).
Curiously, a much weaker $\displaystyle A(n) \lt\frac{1}{\sqrt{n}}$ is still resistant to the inductive argument, whereas a stronger version $\displaystyle A(n) \lt\frac{1}{\sqrt{n + 1}}$ goes through without a hitch.
where mathematical induction applies easily to a stronger inequality and does not seem to work for a weaker one.)
References
A. Engel, , Springer Verlag, 1998, p. 180
D. Fomin,S. Genkin,I. Itenberg, , AMS, 1996, p. 90
S. Savchev, T. Andreescu, , MAA, 2003, p. 51
D. O. Shklyarsky, N. N. Chentsov, I. M. Yaglom, Selected Problems and Theorems of Elementary Mathematics, v 1, Moscow, 1959. (In Russian)
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3和4,1/3-1/4=1/12
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display: 'inlay-fix'1&#47;2*4+1&#47;3*5+1&#47;4*6+1&#47;5*7+...+1&#47;8*10+1&#47;9*11(用简便方法计算)_百度知道
1&#47;2*4+1&#47;3*5+1&#47;4*6+1&#47;5*7+...+1&#47;8*10+1&#47;9*11(用简便方法计算)
二中的寒假作业各位同学帮个忙啊
9*11=(1/4+1/3-1&#47.;2*4+1/6+1&#471/2+1/8-1/5-1&#47.+1/165÷2=53/3*5+1/7+……+1/5+1/8*10+1/10+1/9-1/4*6+1/10-1/5*7+;3-1/11）÷2=106/4-1/2-1&#47.;11）÷2=（1&#47
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2-1&#471/10=9&#47.+1/9*10=1/2*3+1&#47..+1&#47.;2+1/2+1/4+.;3*4+1/2+1/4*5+;10=1/2-1/3-1/3+1&#47.;9-1&#47
原式=(1/2)*（1/2-1/4+1/3-1/5+1/4-1/6.......+1/8-1/10+1/9-1/11)=(1/2)*(1/2+1/3-1/10-1/11)
应该是这样，就写到这了。。后面自己能搞定的把
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出门在外也不愁1&#47;8+10分之1+1&#47;6的和是多少用分数表示是多少_百度知道
1&#47;8+10分之1+1&#47;6的和是多少用分数表示是多少
1&#47;8+10分之1+1&#47;6的和是多少用分数表示是多少
120+20/6=15/120+12&#47解：1/10+1/120=（15+12+20）/120=47/8+1&#47
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