TITLE: Why does $\lim_{k\to\infty} \sqrt[k] {\big | \frac{k^{1/k}-1}{2^k}\big |} = 1/2$?
QUESTION [5 upvotes]: $$\lim_{k\to\infty} \sqrt[k] {\left | \frac{k^{1/k}-1}{2^k}\right |} = \frac12$$ according to a solution I have, but I get the limit to equal zero.
How come it goes to $\frac12$?

REPLY [3 votes]: Since
$$
1+x\le e^x\le\overbrace{1+\frac{x}{1-x}}^{\large\frac1{1-x}}
$$
and $k^{1/k}=e^{\log(k)/k}$, we have
$$
\frac{\log(k)}k\le k^{1/k}-1\le\frac{\log(k)}{k-\log(k)}
$$
Therefore,
$$
\frac12\left(\frac{\log(k)}k\right)^{1/k}\le\left(\frac{k^{1/k}-1}{2^k}\right)^{1/k}\le\frac12\left(\frac{\log(k)}{k-\log(k)}\right)^{1/k}
$$
For $x\ge e$, $\frac1x\le\frac{\log(x)}x\le\frac1e$; thus,
$$
\frac12\left(\frac1k\right)^{1/k}\le\left(\frac{k^{1/k}-1}{2^k}\right)^{1/k}\le\frac12\left(\frac1{e-1}\right)^{1/k}
$$
Thus, by the Squeeze Theorem,
$$
\lim_{k\to\infty}\left(\frac{k^{1/k}-1}{2^k}\right)^{1/k}=\frac12
$$