Claim: $\lim{C^{1/n}}=1$ for $C>0$

Working: $|C^{1/n}-1| < \epsilon$

$ \implies C^{1/n} < \epsilon +1$

$\implies {(1/n)}\ln{C} < \ln{(\epsilon +1)}$

$ \implies n> {\ln{C}/\ln{(\epsilon +1)}}$


Let $ \epsilon >0$ be given.

Choose $N>{\ln{C}/\ln{(\epsilon +1)}}$

Then for any $n>N$, this implies that $|C^{1/n} -1|<\epsilon$

Hence, $\lim{C^{1/n}} =1$ for all $C>0$

Can anyone please verify this proof?

  • $\begingroup$ looks good to me $\endgroup$ – Vasya Mar 7 '18 at 17:50
  • $\begingroup$ The last implication, in the ‘Working‘ part, is true only if $C>1$. It doesn't work if $0<C<1$ because then $\ln C<0$. $\endgroup$ – Bernard Mar 7 '18 at 17:50
  • $\begingroup$ I arrive at $n<ln(ϵ+1)/lnC$ considering the case $0<C<1$, I don't know how to proceed further with this sort of a value $\endgroup$ – A.Asad Mar 7 '18 at 17:53
  • $\begingroup$ if 0<C<1 then take $D=1/C$, expression inside absolute value will be negative so you'll have $1-D^{-1/n}<\epsilon$ and $\ln(1-\epsilon)<-1/n\ln D$, $n>-\ln D/ln(1-\epsilon)$ $\endgroup$ – Vasya Mar 7 '18 at 18:09
  • $\begingroup$ I need it for $0<C<1$, not for C<0 as C>0 (as given by the question) $\endgroup$ – A.Asad Mar 7 '18 at 18:10

It is not correct. For one, you did away with the absolute value.

Try to split it into cases. $C=1$ is obvious, but consider $C>1$. Afterwards, a simple trick takes care of the case $C<1$.

If you want to avoiding logarithms, consider the following. Let $C>1$, so that $C^{1/n}>1$ for all $n\geq 1$. For each $n$, write $C^{1/n}= 1+ d_n$ and notice $d_n > 0$. Then, for each $n$, we have by Bernoulli's ineuqality that

$$C = {\left(C^{1/n}\right)}^n =(1+d_n)^n \geq 1+nd_n.$$

This implies that $0<d_n \leq \frac{C-1}{n}$ and hence $\lim_{n\to\infty}d_n = 0$ by the squeeze theoerem. It follows that

$$\lim_{n\to\infty} C^{1/n} = \lim_{n\to\infty} 1+ d_n = 1,$$

as desired. Finally, for $0<C<1$, we have

$$C^{1/n} = \frac{1}{\left(\frac1C\right)^{1/n}}.$$

Do you think you can conclude?

  • $\begingroup$ I arrive at $n< \ln{(\epsilon+1)}/\ln{C}$ considering the case $0<C<1$, I don't know how to proceed further with this sort of a value $\endgroup$ – A.Asad Mar 7 '18 at 17:52
  • $\begingroup$ Can you please help? $\endgroup$ – A.Asad Mar 7 '18 at 17:59
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    $\begingroup$ I'm not sure what you're allowed to use. Generally this seem to precede logarithms, so I would go for a proof inspired by Newton's binomial theorem, which is 'more basic'. $\endgroup$ – Fimpellizieri Mar 7 '18 at 18:25
  • $\begingroup$ Although we haven't really reached logarithms in analysis, our instructor has allowed us to assume for now that we know how they work and use them $\endgroup$ – A.Asad Mar 7 '18 at 18:30
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    $\begingroup$ Consider my edit. We don't even need the binomial theorem, good ol' Bernoulli will suffice. $\endgroup$ – Fimpellizieri Mar 7 '18 at 20:09

Here is a simpler proof without having to consider different cases (other than implicitly): \begin{align} \bigl|C^{1/n}-1\bigr|<\varepsilon&\iff 1-ε< C^{1/n} < 1 +ε\iff \ln(1-ε)<\frac 1n\ln C <\ln(1+ε)\\ &\iff (n\ln(1-ε)\color{red}< \ln C)\quad\text{and}\quad (\ln C<n\ln(1+ε))\\ &\iff n\color{red}>\underbrace{\frac{\ln C}{\ln(1-ε)}}_{\text{because }\ln(1-ε)<0}\quad\text{and}\quad n>\frac{\ln C}{\ln(1+ε)}\\ &\iff n>\max\biggl(\frac{\ln C}{\ln(1-ε)},\frac{\ln C}{\ln(1+ε)}\biggr). \end{align} Note that if $C\ne 1$, one of these values is positive, the other is negative.

  • $\begingroup$ Thank you! Do I have to show which one is the maximum out of the two? Also, we are taking the maximum because that will work for all the values of n greater than that number, right? $\endgroup$ – A.Asad Mar 7 '18 at 18:16
  • $\begingroup$ The maximum is obvious if you know whether $C>1$ or $C<1$ (rule of signs). And, yes, we take the maximum because it ensures both inequalities will be fulfilled. $\endgroup$ – Bernard Mar 7 '18 at 18:30

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