showing that $c^{\frac{1}{n}}$ is increasing. 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. 
 A: 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.
A: 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.
A: 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.
