Proving $\lim a^{1/n} = 1$ without logarithms?

Question

I was given this problem:

Show if $a>1$ and $n>1$ ($n$ and $a$ are integers) then, $$\lim_{n\to\infty}a^{\frac{1}{n}}=1.$$

The obvious solution is the following:

Take the logarithm in base $a$ of $a^{\frac{1}{n}}$ and logarithm in base $a$ of 1 and you have,

$\lim_{n\to\infty}\frac{1}{n}=0$.

Which is clearly true. Therefore it is equal to $1$.

But the questions ask if you can prove this without logarithms, any solutions?

Answer 1

The basic idea of an "elementary" proof, which I have from Kenneth A. Ross, Elementary Analysis: The Theory of Calculus, is that it behooves us to consider the quantity $s_n := a^{1/n} - 1$, and show that this tends to zero. [Recommendation: See if this hint is enough!]

I will leave you to prove the lemma that $s_n > 0$ for all $n \geq 1$, i.e., if $a > 1$, then $a^{1/n} > 1$ [Proof by contrapositive is the easiest.]

Proof (essentially, Ross): If $s_n = a^{1/n} - 1$, then $a^{1/n} = 1 + s_n$. Raise both sides to the $n$th power and use the binomial formula, and the positivity of $s_n$: For $n \geq 3$,

\begin{align} a &= (a^{1/n})^n && = (1 + s_n)^{n}\\ &&& = \sum_{k = 0}^n \binom{n}{k} s_n^k\\ &&& \geq 1 + n s_n. \end{align}

Thus, $\displaystyle s_n \leq \frac{a-1}{n}$. Use the Squeeze Theorem.

[I seem to remember that an argument in this direction works even for $0 < a < 1$, but since $s_n$ is negative in this case (though still $> -1$) you have to use the $k = 2$ term of the binomial series and watch your error bounds a little more.]

Answer 2

Bernoulli's Inequality says that for $x\ge-n$, $$ 1+x\le\left(1+\frac xn\right)^n\tag{1} $$ For $x\ge-1$, $(1)$ is equivalent to $$ (1+x)^{1/n}\le1+\frac xn\tag{2} $$ Taking reciprocals, we get $$ \left(1-\frac x{1+x}\right)^{1/n}\ge\frac1{1+\frac xn}\tag{3} $$ Substituting $x\mapsto-\frac x{1+x}$ in $(3)$ yields $$ (1+x)^{1/n}\ge\frac1{1-\frac x{1+x}\frac1n}\tag{4} $$ Putting $(2)$ and $(4)$ together gives $$ \frac1{1-\frac x{1+x}\frac1n}\le(1+x)^{1/n}\le1+\frac xn\tag{5} $$ Taking the limit as $n\to\infty$ and using the Squeeze Theorem, shows that, for $x\gt-1$ $$ \lim_{n\to\infty}(1+x)^{1/n}=1\tag{6} $$