1. ## limit of f(n)^(1/n)

Hi all,

In relation to the root test for series convergence, we're given the identity:

$\displaystyle \lim_{n\to\infty}\sqrt[n]{n} = 1$

$\displaystyle \lim_{n\to\infty}\sqrt[n]{(an^b +c)^d} = 1$

(Where a,b,c,d are real constants.)

but I'm thinking there are only limited situations for which the following is true:
$\displaystyle \lim_{n\to\infty}\sqrt[n]{f(n)} = 1$

(Where f(n) is any function of n, including, for example, f(n) = n!)

like $\displaystyle f(n) = n^n$ in this situation would diverge.

So, what is the truth table for the above equation? It seems to me that the root test would be easier to conduct with more general forms available, re my first equation versus my second.

Any thoughts appreciated.

Thanks,
Brian

2. If $\displaystyle f(n)>0, \forall n$ is...

$\displaystyle \sqrt[n] {f(n)}= e^{\frac {\ln f(n)}{n}}$ (1)

... so that the request...

$\displaystyle \lim_{ n \rightarrow \infty} \sqrt[n] {f(n)}= 1$ (2)

... is equivalent to the request...

$\displaystyle \lim_{ n \rightarrow \infty} \frac {\ln f(n)}{n}= 0$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Thank you ChiSigma.

Hmm.. So now I'm really trying to use that information in a way that produces useful identities for evaluating the root test for convergence. I suppose the question is pedagogical.. In other words, I wonder why in my textbook the section for root test gives the $\displaystyle \lim_{n\to\infty}\sqrt[n]{n} = 1$ identity, when it just as easily could have given $\displaystyle \lim_{n\to\infty}\sqrt[n]{n^k} = 1$, or an even more complex form.

From what you've given me, it seems clear that if $\displaystyle f(n) = g(n)^n$ then $\displaystyle \lim_{ n\to\infty} \frac {\ln f(n)}{n}$ diverges (assuming g(n) does not contain the power 1/n, I just don't know how to better express my intent here,) but I am not sure what else I can extract from that.

I guess I really just need to do a lot more root test problems so that I can identify real examples of when more complex identities would be useful.

Brian