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Let $x_n>0 \space \forall \space n \in N$

If $\lim_{n\to\infty} \frac{x_{n+1}}{x_n}=L$, then $\lim_{n\to\infty} (x_n)^{1/n}=L$

This theorem is given in my book without proof. It looks very strange to me. Any suggestion about how to prove or derive it?

3 Answers3

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Suppose $\lim_\limits{n\to\infty}\frac{x_{n+1}}{x_n}=L$. Then $$\frac{1}{L}=\frac1{\lim_{n\to\infty}\frac{x_{n+1}}{x_n}}=\lim_{n\to\infty}\frac{x_n}{x_{n+1}}$$ by continuity of $\frac1x$ for $x>0$. We recognize $\lim_\limits{n\to\infty}\frac{x_n}{x_{n+1}}$ as the radius of convergence for the power series $$\sum_{n=0}^\infty x_nx^n$$ which we will denote $R$. Similarly, we know that the radius of convergence for the above series is given as $\frac1{\limsup_\limits{n\to\infty}\sqrt[n]{x_n}}$ since $x_n>0$. Notice that $$\frac1L=R=\frac1{\limsup_\limits{n\to\infty}\sqrt[n]{x_n}}=\frac1{\lim_\limits{n\to\infty}\sqrt[n]{x_n}}$$ in this case (due to our original assumption).

Dave
  • 13,568
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Here is a proof from first principles.

If $\lim_{n\to\infty} \frac{x_{n+1}}{x_n}=L $, then, for any $c > 0$, for all $n > n_0(c)$, $L-c \lt \frac{x_{n+1}}{x_n} \lt L+c $.

Therefore, for any $N > n_0(c)$, $(L-c)^{N-n_0(c)} \lt \prod_{n=n_0(c)}^{N-1} \frac{x_{n+1}}{x_n} =\dfrac{x_N}{x_{n_0(c)}} \lt (L+c)^{N-n_0(c)} $ so that $x_{n_0(c)}(L-c)^{N-n_0(c)} \lt x_N \lt x_{n_0(c)}(L+c)^{N-n_0(c)} $ or $x_{n_0(c)}\dfrac{(L-c)^{N}}{(L-c)^{n_0(c)}} \lt x_N \lt x_{n_0(c)}\dfrac{(L+c)^{N}}{(L+c)^{n_0(c)}} $.

Taking the $N$th root, and rearranging a little, $(L-c)\left(\dfrac{x_{n_0(c)}}{(L-c)^{n_0(c)}}\right)^{1/N} \lt x_N^{1/N} \lt (L+c)\left(\dfrac{x_{n_0(c)}}{(L+c)^{n_0(c)}}\right)^{1/N} $.

Both $\dfrac{x_{n_0(c)}}{(L-c)^{n_0(c)}}$ and $\dfrac{x_{n_0(c)}}{(L+c)^{n_0(c)}}$ are independent of $N$, so $\lim_{N\to \infty} \left(\dfrac{x_{n_0(c)}}{(L\pm c)^{n_0(c)}}\right)^{1/N} =1 $.

Therefore, for large enough $N$, $1-c \lt \left(\dfrac{x_{n_0(c)}}{(L\pm c)^{n_0(c)}}\right)^{1/N} \lt 1+c $, so that $(L-c)(1-c) \lt x_N^{1/N} \lt (L+c)(1+c) $.

Now let $c \to 0$ and, for each $c$, let $N$ be sufficiently large and we are done.

marty cohen
  • 107,799
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If $x_n>0 ,\forall n \in \mathbb{N}$ then $L>0$.

If $a_{n+1}-a_n \to a \in \mathbb{R}$ then $\frac{a_n}{n} \to a$

This is a corollary of Cesaro's theorem.

$\frac{x_{n+1}}{x_n} \to L$ thus $\log{\frac{x_{n+1}}{x_n}}=\log{x_{n+1}-\log{x_n}} \to \log{L}$

So from the corollary we have that $$\frac{\log{x_n}}{n} \to \log{L} \Rightarrow \log{\sqrt[n]{x_n}} \to \log{L} \Rightarrow \sqrt[n]{x_n} \to L$$

Marios Gretsas
  • 21,906