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An Inequality: |
| (1) | 1/2·3/4·5/6· ... ·99/100 < 1/10. |
Copyright © 1996-2009 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),
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whose proof is a slight modification of the above with A and B defined as
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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).
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(3) supplies an edifying curiosity. By itself, it is easily proven by mathematical induction. However, its weakened version
| (3') |
A(n) < 1/(3n),
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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).
Copyright © 1996-2009 Alexander Bogomolny
To remind,
| A(n) = 1/2·3/4·5/6· ... ·(2n-1)/2n |
and we wish to prove (3n+1).
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A(1) = 1/2 = 1/(3·1+1).
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But already for n = 2,
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A(2) = 1/2·3/4 = 3/8 < 1/7 = 1/(3·2+1),
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because upon squaring 9/64 < 1/7, for
| (4) |
A(k) < 1/(3k+1).
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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).
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Since A(k+1) = A(k)·(2k+1)/(2k+2), (4) implies
| (6) |
A(k+1) < (2k+1)/(2k+2)·1/(3k+1).
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Now square the right hand side in (6):
| (7) |
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which is exactly the right-hand side of (5) and proves (6).
(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)
Copyright © 1996-2009 Alexander Bogomolny
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