$N^{th}$ root of a sequence converges to the same limit.

Question

I need to prove that for a sequence $\{a_n\}$ of positive numbers such that $\lim\limits_{n \to \infty} a_n = L$. Show $\lim_{n \to \infty} \sqrt[n]{(a_1a_2\ldots a_n)}= L$. Can someone help me with this.

Answer 1

Hint: Let $\epsilon>0.$ Then there exists $N$ such that $a_n<L+\epsilon$ for $n>N.$ Thus

$$(a_1a_2\dots a_n)^{1/n}= (a_1a_2\dots a_N)^{1/n}(L+\epsilon)^{(n-N)/n}.$$

Take the $\limsup$ as $n\to \infty$ to see

$$\limsup (a_1a_2\dots a_n)^{1/n}\le 1\cdot (L+\epsilon).$$

This shows $\limsup(a_1a_2\dots a_n)^{1/n}\le L.$

Answer 2

The only limit needed is that $\lim_{n \to \infty} a^{1/n} = 1 $ for any $a > 0$. Here's a proof of that from "What is Mathematics" that only uses Bernoulli's inequality $(1+x)^n \ge 1+nx$ for $x > 0, n \ge 1$.

If $a > 1$ let $b = a-1$ so $b \gt 0$ and $a^{1/n} = 1+c$. Then $a =1+b =(1+c)^n \ge 1+nc \gt nc $ so $c < \dfrac{a}{n}$ so $a^{1/n}-1 \lt \dfrac{a}{n} $.

Similarly, if $0 < a < 1$, then $a = \dfrac1{1+b}$ where $b = \dfrac1{a}-1 > 0$. If $1+c = (1+b)^{1/n}$ then $1+b =(1+c)^n \ge 1+nc \gt nc $ so $c \lt \dfrac{1+b}{n} =\dfrac1{an} $ so $a^{1/n} =\dfrac1{(1+b)^{1/n}} =\dfrac1{1+c} $ and $1-a^{1/n} =1-\dfrac1{1+c} \lt 1-\dfrac1{1+\dfrac1{an}} = 1-\dfrac{an}{an+1} = \dfrac{1}{an+1} $.

In both cases $a^{1/n}-1 \to 0$.

Now, the main attraction.

Let $A_n =\prod_{k=1}^n a_k$.

If $L = 0$, for any $c > 0$ there is a $n(0)$ such that $|a_n| < c$ for $n \ge n(0)$. Then, for $n > n(0)$,

$\begin{array}\\ |A_n| &=|\prod_{k=1}^n a_k|\\ &=\prod_{k=1}^n |a_k|\\ &=\prod_{k=1}^{n(0)} |a_k|\prod_{k=n(0)+1}^n |a_k|\\ &\le\prod_{k=1}^{n(0)} |a_k|\prod_{k=n(0)+1}^n c\\ &=\prod_{k=1}^{n(0)} |a_k|c^{n-n(0)}\\ &=\dfrac{\prod_{k=1}^{n(0)} |a_k|}{c^{n(0)}}c^{n}\\ &=p(c)c^{n} \qquad p(c) = \dfrac{\prod_{k=1}^{n(0)} |a_k|}{c^{n(0)}}\\ \text{so}\\ |A_n|^{1/n} &\le (p(c))^{1/n}c\\ &\le 2c \qquad\text{for large enough } n\\ \end{array} $

Therefore, for any $c > 0$ $|A_n|^{1/n} \le 2c$ for all large enough $n$, so $\lim_{n \to \infty} |A_n|^{1/n} =0$.

For $L > 0$ we do the same decomposition of the product but with different bounds.

If $L > 0$ then, for any $c > 0$ such that $c < L/2$, there is a $n(c)$ such that $L-c < a_n < L+c$ for $n > n(c)$, or $1-\frac{c}{L} < \frac{a_n}{L} < 1+\frac{c}{L}$ or $1-d < \frac{a_n}{L} < 1+d$ where $d = \frac{c}{L} $.

Then, for $n > n(0)$,

$\begin{array}\\ |A_n| &=|\prod_{k=1}^n a_k|\\ &=\prod_{k=1}^n |a_k|\\ &=\prod_{k=1}^{n(0)} |a_k|\prod_{k=n(0)+1}^n |a_k|\\ \text{so}\\ \dfrac{|A_n|}{L^n} &=\dfrac{\prod_{k=1}^{n(0)} |a_k|\prod_{k=n(0)+1}^n |a_k|}{L^n}\\ &=\dfrac{\prod_{k=1}^{n(0)} |a_k|}{L^{n(0)}}\dfrac{\prod_{k=n(0)+1}^n |a_k|}{L^{n-n(0)}}\\ &=p(c)\prod_{k=n(0)+1}^n \dfrac{|a_k|}{L} \qquad p(c) = \dfrac{\prod_{k=1}^{n(0)} |a_k|}{L^{n(0)}}\\ &\text{Now we get upper and lower bounds}\\ \dfrac{|A_n|}{L^n} &=p(c)\prod_{k=n(0)+1}^n \dfrac{|a_k|}{L}\\ &\lt p(c)\prod_{k=n(0)+1}^n (1+d)\\ &= p(c)(1+d)^{n-n(0)}\\ &= \dfrac{p(c)}{(1+d)^{n(0)}}(1+d)^{n}\\ &= q(c)(1+d)^{n} \qquad q(c) = \dfrac{p(c)}{(1+d)^{n(0)}}\\ \text{so}\\ \dfrac{|A_n|^{1/n}}{L} &< \sqrt[n]{q(c)}(1+d)\\ &< (1+d)(1+d) \qquad \text{for large enough } n\\ &=(1+d)^2\\ \end{array} $

Since we can choose any $d > 0$ by making $c = Ld$, $\lim\sup \dfrac{|A_n|^{1/n}}{L} \le 1 $.

Similarly, using the lower bound for $\dfrac{a_n}{L}$, we can show that $\lim\inf \dfrac{|A_n|^{1/n}}{L} \ge 1 $.

Therefore $\lim_{n \to \infty}= \dfrac{|A_n|^{1/n}}{L} = 1 $.