An Inequality:
1 + 2^{-2} + 3^{-2} + 4^{-2} + ... + n^{-2} < 2
Prove the following inequality for all integer n greater than 1:
(1) | 1 + 2^{-2} + 3^{-2} + 4^{-2} + ... + n^{-2} < 2 |
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Copyright © 1996-2018 Alexander Bogomolny
(This page has been inspired by a post from Torsten Hennig at the Mathforum.)
Mathematical induction is a reasonable method to apply to proving (1) "for all n". As is the case with another example and yet another one, there does not appear to be an obvious way to make induction work for (1).
(1) | 1 + 2^{-2} + 3^{-2} + 4^{-2} + ... + n^{-2} < 2 |
While it is easy to compare and grow the partial sums on the left, the right hand side is static. We can make it dynamic by strengthening the inequality:
(2) | 1 + 2^{-2} + 3^{-2} + 4^{-2} + ... + n^{-2} < 2 - 1/n. |
The latter is easily amenable to an inductive argument. Denote the left-hand side of (2) as A(n). The verification of (2) for
A(k) < 2 - 1/k
For n = k + 1, we have
A(k+1) | = A(k) + (k + 1)^{-2} |
< (2 - 1/k) + 1/(k + 1)^{2} | |
< 2 - 1/k + 1/[k(k + 1)] | |
= 2 - 1/k·(1 - 1/(k + 1)) | |
= 2 - 1/k · k/(k + 1) | |
= 2 - 1/(k + 1), |
which is exactly (2) for
If we pass to the limit as n grows, A(n) will become the infinite sum in the left hand side of (1), the right hand side will become 2. However, the inequlaity will change slightly:
(1') | 1 + 2^{-2} + 3^{-2} + 4^{-2} + ... ≤ 2 |
What do we do with the possibility of "="?
Those who know that
(1'') | 1 + 2^{-2} + 3^{-2} + 4^{-2} + ... ≤ 1.8 (< 2) |
It is easy to see that all stages of the inductive proof go through for (1'').
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Copyright © 1996-2018 Alexander Bogomolny