The exercise asks me to show that lim$(c^{\frac{1}{n}}) = 1 ,where, 0<c<1$, using subsequences.
Please verify my proof and help with the last part.
Here:
$$\frac{1}{n+1} < \frac{1}{n} \Rightarrow c^{\frac{1}{n+1}} < c^{\frac{1}{n}} $$ Hence, it is a monotone decreasing sequence.
Since, $ 0<c<1$, we find, $0<c^{\frac{1}{n}}<1 $ Therefore, the sequence is bounded.
By the Monotone Convergence Theorem, the sequence is convergent.
Let $lim (c^{\frac{1}{n}}) = x $
Choosing $n*=2n$ We get a subsequence, $c^{\frac{1}{n*}} = c^{\frac{1}{2n}} = (c^{\frac{1}{n}})^{\frac{1}{2}} $
As limit of a sequence is equal to the limit of its subsequence, $$ x = lim (c^{\frac{1}{2n}}) = (\lim\limits (c^{\frac{1}{n}}))^{\frac{1}{2}} = x^{\frac{1}{2}} $$ $$ \Rightarrow x^{2} - x = 0 \Rightarrow x = 0, 1 $$ At this point in the proof, i am not sure how to proceed to show that x cannot be = 0. Or if there is something wrong with my proof, which is why i have reached such a result. Thank you!
$(c^{1/n})$ is actually increasing, not decreasing. [For $0<c<1$ $x<y$ imples $c^{x} >c^{y}$], Hence the limit cannot be $0$.