2
$\begingroup$

Problem: Show that if $0<c<1$ then lim $c^{\frac{1}{n}}=1$ by the monotone convergence theorem. I already have that this sequence is bounded above by 1. But to show that $c_n=c^{\frac{1}{n}}$ is increasing is what I'm having trouble with. I know im supposed to use induction.

$\endgroup$
  • $\begingroup$ A deceasing sequence that is bounded below by 1. My book proves that already but they don't show how to prove its decreasing when x>1. $\endgroup$ – user60887 Oct 30 '13 at 17:52
  • $\begingroup$ You already have that $c^{\frac1n}$ is bounded, so is $c^{\frac1{n(n+1}}$. (hint) $\endgroup$ – Carlos Eugenio Thompson Pinzón Oct 30 '13 at 17:53
  • $\begingroup$ So you mean if $\frac{c_{n+1}}{c_n}$ is greater then 0 so its increasing? $\endgroup$ – user60887 Oct 30 '13 at 17:59
  • $\begingroup$ @CarlosEugenioThompsonPinzón Check your arithmetic. $\endgroup$ – Don Larynx Oct 30 '13 at 18:01
  • 2
    $\begingroup$ @DonLarynx Remember we are trying to prove that the sequence is increasing (for $0<c<1$). So we know that $c^k$ is bounded (and positive), particularly $c^{\frac1{n(n+1)}}<1$ Multiply by positive $c^{\frac1{n+1}}$ at both sides. What do you get? $\endgroup$ – Carlos Eugenio Thompson Pinzón Oct 30 '13 at 18:10
2
$\begingroup$

You already know that $0<c^{\frac1k}<1$ for any $k$ when $0<c<1$.

This means that (for $k=n(n+1)\,$).\begin{align} c^{\frac1{n(n+1)}}&<1&&\text{then mlutiply by possitive $c^{\frac1{n+1}}$} \\ c^{\frac1{n+1}}\cdot c^{\frac1{n(n+1)}}&<c^{\frac1{n+1}} \\ c^{\frac1{n+1}+\frac1{n(n+1)}}&<c^{\frac1{n+1}} \\ c^{\frac1{n}}&<c^{\frac1{n+1}}. \end{align} Therefor $c_n=c^{\frac1n}$ is increasing.

$\endgroup$
1
$\begingroup$

Suppose $0 < c < 1$. Consider the set $C$ in a metric $X, d$. We show $c^{\frac{1}{n}}$ is increasing.

For $n = 1$, we have $c^{\frac{1}{n}} > c^{\frac{1}{n+1}}$, or $c^{{\frac{1}{n} - \frac{1}{n+1}}} = c^k > 1$. Since the exponential expression is always positive, and since the exponential term also converges to $0$ for really large $n$ (it's clear using limits) we have $\lim_{k\to\infty}c^k = 1$. Since $c > 0$, then $k$ must converge to $0$. But since $c < 1$ then as $k$ gets smaller, $c^k$ gets bigger.

$\endgroup$
0
$\begingroup$

I figured out another way. Let $c_n=c^\frac{1}{n}$ then $\frac{c_{k+1}}{c_{k}}=\frac{c^\frac{1}{k+1}}{c^\frac{1}{k}}=c^{\frac{1}{k+1}-\frac{1}{k}}=c^\frac{-1}{k(k+1)}$ But since $c<1$ it follows $c^\frac{-1}{k(k+1)}=\frac{1}{c^\frac{1}{k(k+1)}}>1$. Thus $c_n$ is increasing.

$\endgroup$

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.