3
$\begingroup$

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 $x_{n+1}<x_{n+2}$?

$\endgroup$
2
$\begingroup$

$c < 1$ 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 $c = {c^{\frac{1}{n + 1}}}^{n+1} \le {c^{\frac{1}{n}}}^{n+1} = c*c^{\frac{1}{n}} < c$

$\endgroup$
2
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.