An Inequality: |
| (1) | 1/2·3/4·5/6· ... ·99/100 < 1/10. |
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Copyright © 1996-2012 Alexander Bogomolny
Denote the left-hand side of the inequality A:
A = 1/2·3/4·5/6· ... ·99/100.
And introduce its nemesis B:
B = 2/3·4/5·6/7· ... ·98/99.
Factor by factor, the fractions in B exceed those in A:
2/3 > 1/2, 4/5 > 3/4, ..., 98/99 > 97/98, 1 > 99/100.
From this it follows that A < B. Note that, due to the choice of B, in the product AB most of the terms cancel out:
A2 < AB = 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
| (2) | 1/2·3/4·5/6· ... ·(2n-1)/2n < 1/√2n, |
whose proof is a slight modification of the above with A and B defined as
A(n) = 1/2·3/4·5/6· ... ·(2n-1)/2n,
B(n) = 2/3·4/5·6/7· ... ·(2n-2)/(2n-1).
As we shall see shortly, (1) and (2) are quite weak: A(n) has a much better bound, viz.
| (3) | A(n) < 1/√3n+1. |
(3) supplies an edifying curiosity. By itself, it is easily proved by mathematical induction. However, its weakened version
| (3') | A(n) < 1/√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 proof for (3).
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Copyright © 1996-2012 Alexander Bogomolny
To remind,
A(n) = 1/2·3/4·5/6· ... ·(2n-1)/2n
and we wish to prove
A(1) = 1/2 = 1/√3·1+1.
But already for n = 2,
A(2) = 1/2·3/4 = 3/8 < 1/√7 = 1/√3·2+1,
because upon squaring 9/64 < 1/7, for
| (4) | A(k) < 1/√3k+1. |
We are going to prove that, for n = k+1, (3) also holds
| (5) | A(k+1) < 1/√3(k+1)+1 = 1/√3k+4. |
Since A(k+1) = A(k)·(2k+1)/(2k+2), (4) implies
| (6) | A(k+1) < (2k+1)/(2k+2)·1/√3k+1. |
Now square the right hand side in (6):
| (7) |
|
which is exactly the right-hand side of (5) and proves (6).
Curiously, a much weaker A(n) < 1/√n is still resistant to the inductive argument, whereas a stronger version
(There is another example where mathematical induction applies easily to a stronger inequality and does not seem to work for a weaker one.)
References
- D. Fomin,S. Genkin,I. Itenberg, Mathematical Circles (Russian Experience), AMS, 1996, p. 90
- S. Savchev, T. Andreescu, Mathematical Miniatures, 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)
Inequalities to prove:
- An Inequality for Grade 8
- An Extension of the AM-GM Inequality
- A Mathematical Rabbit out of an Algebraic Hat
- An Inequality With an Infinite Series
- An Inequality: 1/2 * 3/4 * 5/6 * ... * 99/100 less than 1/10
- A Low Bound for 1/2 * 3/4 * 5/6 * ... * (2n-1)/2n
- An Inequality: Easier to prove a subtler inequality
- Inequality with Logarithms
- An inequality: 1 + 1/4 + 1/9 + ... less than 2
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Copyright © 1996-2012 Alexander Bogomolny
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