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Show that the sequence $(x_n)=c^{\frac{1}{n}}$ is increasing for $0 < c < 1$.

I am trying to do this using induction. We see for the base case that $x_1 = c$ and $x_2 = c^{0.5}$, so clearly $x_1 < x_2$.

For the induction step, we assume $x_n < x_{n+1} \Leftrightarrow c^{\frac{1}{n}} < c^{\frac{1}{n+1}}$.

This is where I am stuck. How can we show that our assumption implies that $x_{n+1}<x_{n+2}$?

flubsy
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2 Answers2

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One way to show that $x_n=c^{1/n}$ is increasing is to analyze the ratio $x_{n+1}/x_n$. To that end, we write

$$\frac{x_{n+1}}{x_n}=\frac{c^{1/n+1}}{c^{1/n}}=\frac{1}{c^{1/(n(n+1))}}$$

Inasmuch as $0<c<1$, we instantly see that $x_{n+1}/x_n\ge 1$ and we are done!

Another way forward is to realize that $x_n<1$ for all $n$. Then, since $0<(x_n)^n=(x_{n+1})^{n+1}=c<1$, we have

$$x_{n+1}=(x_n)^{n/(n+1)}>x_n$$

and we are done!

Mark Viola
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$c < 1$ so $c^n > c^n*c = c^{n + 1}$

$c^{\frac{1}{n}} < 1$ for all positive n. (Otherwise $c = (c^{\frac{1}{2}})^n \ge 1.)$

So $c^{\frac{1}{n}} < c^{\frac{1}{n + 1}}$. Other wise $c^{\frac{1}{n + 1}} \le c^{\frac{1}{n}}$ would imply $c = {c^{\frac{1}{n + 1}}}^{n+1} \le {c^{\frac{1}{n}}}^{n+1} = c*c^{\frac{1}{n}} < c$

fleablood
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