0
$\begingroup$

I want to find the radius of convergence for $\sum n^k z^n$, where $k \in \mathbb{N}$. Using the root test, I want to find $$lim_{n \rightarrow \infty} \; n^{k/n}$$

I know $lim_{n \rightarrow \infty} \; n^{1/n} = 1$, so using properties of convergent sequences, $lim_{n \rightarrow \infty} \; n^{k/n} = 1^k = 1 $.


First, is this right/Is there a better way to prove this?

Second, this makes no intuitive sense, shouldn't this sequence diverge since $n^{k/n} \geq n^{1/n}$?

$\endgroup$
0
$\begingroup$

Yes, it is correct. And, since $\lim_{n\to\infty}n^{1/n}=1$, the fact that you always have $n^{k/n}\geqslant n^{1/n}$ does not imply at all that $(n^{k/n})_{n\in\mathbb N}$ diverges.

$\endgroup$

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.