# 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

Solution

(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 n = 2 is immediate. Assume (2) holds for n = k:

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 n = k + 1. So (2) holds for all n > 1.

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 Σn-2 = π2 / 6, will realize that A(n) could not be close to 2, because π2 / 6 is less than 1.65, let alone 2. Note that the inductive step only deals with the variable part -1/k. The fixed part, 2, is carried over without change. So, perhaps, we can start with a quantity smaller than 2. We can indeed, for A(2) = 1 + 1/4 = 1.25 < 1.8 - 1/2 say. Thus instead of (1') we can try proving

 (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'').