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 a_{1}, a_{2}, … a_{n},

(a_{1} + a_{2} + …. + a_{n})/n ≥(a_{1}a_{2} … a_{n})^{1/n}

with equality if and only if a_{1} = a_{2} = a_{3} … = a_{n}.

**Proof**

P(1) is trivially true, i.e. a_{1}/1 = a_{1}^{1/1}.

P(2) If a_{1} and a_{2} are ≥0 then it is easily shown that …

(a_{1} + a_{2})^{2}≥ 4a_{1}a_{2} and so (a_{1} + a_{2})/2 ≥ √(a_{1}a_{2}).

If (a_{1}+ a_{2})/2 = √(a_{1}a_{2}) then (a_{1} – a_{2})^{2} = 0 and so a_{1} = a_{2}.

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

Suppose that a_{1}, a_{2}, … a_{n} are n non-negative numbers. We can assume that a_{1}≤a_{2}≤a_{3} …≤ a_{n}. If not then we can choose a re-arrangement of a_{1}, a_{2}, … a_{n} that does have this property. It now follows that …

(a_{2} + a_{3} + … a_{n})/(n-1)≥(a_{2}a_{3} … a_{n})^{1/(n-1)}or

a_{2}a_{3}… a_{n}≤((a_{2}+ a_{3}+ … a_{n})/(n-1))^{(n-1)} with equality iff a_{2} = a_{3} = a_{4} … = a_{n}.

Now let D = (a_{2} – a_{1}) + (a_{3} – a_{1}) + … + (a_{n} – a_{1}). Clearly D≥0 with equality iff a_{1}=a_{2}=a_{3} … = a_{n}.

D = a_{2} + a_{3} + … +a_{n} – (n-1)a_{1} and so, a_{1} + D/(n-1) = (a_{2} + a_{3} + … + a_{n})/(n-1).

So, a_{2}a_{3}… a_{n}≤(a_{1} + D/(n-1))^{(n-1)} with equality iff a_{2}= a_{3}= a_{4}… = a_{n}.

Hence, multiplying by a_{1} we get a_{1}a_{2}a_{3}… a_{n}≤a_{1}(a_{1}+ D/(n-1))^{(n-1)}.

Applying (*) above we get a_{1}a_{2}a_{3}… a_{n}≤a_{1}(a_{1}+ D/(n-1))^{(n-1)}≤(a_{1}+ D/n)^{n}, or

a_{1}a_{2}a_{3}… a_{n}≤(a_{1}+ D/n)^{n}=((a_{1} + a_{2} + … a_{n})/n)^{n}, with equality iff D=0 or iff a_{1}=a_{2}=a_{3}… = a_{n}.

Hence it follows that (a_{1} + a_{2} + … +a_{n})/n≥ ^{n}√(a_{1}a_{2} … a_{n}) with equality iff a_{1}=a_{2}=a_{3}… = 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)}– n^{n}> (n+1)^{n}-n^{n }and from this it follows that …

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

Simplifying n^{n}(n-1)>n(n(n+1))^{(n-1)/2}gives 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**.

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