$\sqrt[\infty]{\infty^2}$ in limit of series using root test I'm trying to solve a problem to show if the infinite series $\sum\limits_{k=1}^{\infty}\dfrac{k^2}{2^k}$ converges or diverges using the root test.
When put in limit form, I got $\lim\limits_{k\to\infty}\sqrt[k]{\bigg|\dfrac{k^2}{2^k}\bigg|}$
and then I simplified that to $\lim\limits_{k\to\infty}\dfrac{\sqrt[k]{k^2}}{2}$.
Now I'm not entirly sure what to do with this.  If you plug in infinity, it would give you $\dfrac{\sqrt[\infty]{\infty^2}}{2}$ but I've never seen the "infinite root" of something.  If I plug $\sqrt[\infty]{\infty^2}$ into my calculator (TI nspire) it says the answer is 1, but when I plug it into Wolfram Alpha, it says the answer is infinity.  Using the root test, this would either show that the series converges or diverges, respectively.  I'm not sure which one to trust, and I'm not even sure why $\sqrt[\infty]{\infty^2}$ would be 1 or $\infty$.  Could anyone explain that to me?
 A: What you call an "infinite root" is known to limit to $1$:
$$\lim_{x\to\infty}x^{1/x}=1$$
You can take it for granted, or you can prove it by writing
$$x^{1/x}=e^{\ln x/x}$$
and observing that $x$ grows faster than $\ln x$ for large $x$, making the exponent go to $0$ at infinity.
The square in $k^2$ doesn't do much in this case: because $k>0$, it doesn't affect the sign, and you can write the expression as $(\sqrt[k]{k})^2$.
Also, you can't just write it as $\sqrt[\infty]{\infty}$. This expression only makes sense as a limit, you can't compute with infinities if you don't know how fast and in what way they get to infinity.
A: As an arithmetic operation on the extended reals, it is left undefined, because there is no way to continuously extend $\sqrt[x]{y}$ to have a value there: e.g. for a fixed $y > 0$, $\sqrt[x]{y} \to 1$ as $x \to \infty$, but for a fixed $x > 0$, $\sqrt[x]{y} \to \infty$ as $y \to \infty$.
Correspondingly, when viewed as a limit form rather than an arithmetic operation, $\sqrt[\infty]{\infty}$ is an indeterminate form.
Thus, more work has to be done before you can obtain the limit. As the other answer points out, $x^{1/x} \to 1$ is the key fact: it is actually one of the most important facts to be able to use fluently in order to use the root test efficiently.
