The following result is essential to the proof.

If xεR, x≥0, then (1+x/(n-1))^n-1≤(1+x/n)ⁿ, nεω, n≥2. Equality holds iff x=0. (*)

Proof

If xεR, x≥0, and nεω, n≥2, then the following expression may be expanded using the binomial Theorem …

(1+x/n)ⁿ-(1+x/(n-1))^n-1

= 1 +∑_1≤r≤n n(n-1) … (n-r+1)/r! (x/n)^r

-(1 +∑_1≤r≤n-1 (n-1)(n-2) … (n-r)/r! (x/(n-1))^r)

=∑_1≤r≤n-1 x^r/r![(n-1)/n … (n-r+1)/n – (n-2)/(n-1) … (n-r)/(n-1)]

+ (x/n)ⁿ.

It is straightforward to show that …

[(n-1)/n … (n-r+1)/n – (n-2)/(n-1) … (n-r)/(n-1)]>0 since, (n-1)/n > (n-2)/(n-1) … and (n-r+1)/n > (n-r)/(n-1).

Hence it follows that (1+x/n)ⁿ-(1+x/(n-1))^n-1≥(x/n)ⁿ if x≥0.

The result, i.e. (1+x/(n-1))^n-1≤(1+x/n)ⁿ, follows easily from this, and if equality holds, then (x/n)ⁿ = 0, and so x=0.

Statement and proof of the Arithmetic-Geometric Mean inequality.

For each nεω, let P(n) be the proposition that for any set of non-negative numbers a₁, a₂, … a_n,

(a₁ + a₂ + …. + a_n)/n ≥(a₁a₂ … a_n)^1/n

with equality if and only if a₁ = a₂ = a₃ … = a_n.

Proof

P(1) is trivially true, i.e. a₁/1 = a₁^1/1.

P(2) If a₁ and a₂ are ≥0 then it is easily shown that …

(a₁ + a₂)²≥ 4a₁a₂ and so (a₁ + a₂)/2 ≥ √(a₁a₂).

If (a₁+ a₂)/2 = √(a₁a₂) then (a₁ – a₂)² = 0 and so a₁ = a₂.

Now suppose that P(n-1) is true for n-1 ε ω and n – 1≥ 2.

Suppose that a₁, a₂, … a_n are n non-negative numbers. We can assume that a₁≤a₂≤a₃ …≤ a_n. If not then we can choose a re-arrangement of a₁, a₂, … a_n that does have this property. It now follows that …

(a₂ + a₃ + … a_n)/(n-1)≥(a₂a₃ … a_n)^1/(n-1)or

a₂a₃… a_n≤((a₂+ a₃+ … a_n)/(n-1))^(n-1) with equality iff a₂ = a₃ = a₄ … = a_n.

Now let D = (a₂ – a₁) + (a₃ – a₁) + … + (a_n – a₁). Clearly D≥0 with equality iff a₁=a₂=a₃ … = a_n.

D = a₂ + a₃ + … +a_n – (n-1)a₁ and so, a₁ + D/(n-1) = (a₂ + a₃ + … + a_n)/(n-1).

So, a₂a₃… a_n≤(a₁ + D/(n-1))^(n-1) with equality iff a₂= a₃= a₄… = a_n.

Hence, multiplying by a₁ we get a₁a₂a₃… a_n≤a₁(a₁+ D/(n-1))^(n-1).

Applying (*) above we get a₁a₂a₃… a_n≤a₁(a₁+ D/(n-1))^(n-1)≤(a₁+ D/n)ⁿ, or

a₁a₂a₃… a_n≤(a₁+ D/n)ⁿ=((a₁ + a₂ + … a_n)/n)ⁿ, with equality iff D=0 or iff a₁=a₂=a₃… = a_n.

Hence it follows that (a₁ + a₂ + … +a_n)/n≥ ⁿ√(a₁a₂ … a_n) with equality iff a₁=a₂=a₃… = a_n. Thus P(n) is true and the result follows by the principle of Mathematical Induction.

Applications of the AGM to the sequence n^1/n, where nεω, n≥1.

Numerical computation of the first few terms of this sequence shows that the first term 1^1/1 = 1, the second is 2^1/2 = √2 and the third term is ³√3 ( the maximum value of the sequence). For n≥3 the sequence appears to be monotone decreasing, tending to a limit of 1.

Firstly, 2 subsequent values cannot be equal. Otherwise n^1/n = (n+1)^1/(n+1) for some n. From this equation (if true) it follows that n⁽ⁿ⁺¹⁾ = (n+1)ⁿ. If n=1 then it follows that 1 = 2 which is obviously false. If n>1 then n and n+1 must have a common prime factor, again false since n and n+1 are coprime.

So, for all nεω, n≥1, either n^1/n> (n+1)^1/(n+1) or n^1/n< (n+1)^1/(n+1).

Suppose that n^1/n> (n+1)^1/(n+1). Then, n⁽ⁿ⁺¹⁾> (n+1)ⁿ.

So, n⁽ⁿ⁺¹⁾– nⁿ> (n+1)ⁿ-nⁿ and from this it follows that …

nⁿ(n-1)> (n+1)^n-1 + (n+1)^n-2n + … + n^n-1 >n(n(n+1))^(n-1)/2 (applying the AGM).

Simplifying nⁿ(n-1)>n(n(n+1))^(n-1)/2gives n-1>((n+1)/n)^(n-1)/2>1. Hence it follows that n≥3.

So, for all nεω, n≥3, n^1/n> (n+1)^1/(n+1).

To prove that lim _n→∞ n^1/n = 1, first observe that n^1/n> 1 for all n> 1, otherwise n^1/n≤1 for some n>1. n^1/n≤ 1 implies that n≤1, which clearly gives a contradiction, so n^1/n> 1 for all n> 1.

Next, if n>2, we apply the AGM inequality to the set of n positive numbers 1, 1, … ,1 (n-2 times),√n,√n.

This gives (n-2+2√n)/n>n^1/n, or 1-2/n+2/√n>n^1/n. This last inequality implies that as n→∞, n^1/n approaches arbitrarily close to 1. Hence the result lim_n→∞ n^1/n= 1.