Why does $\lim_{k\to\infty} \sqrt[k] {\big | \frac{k^{1/k}-1}{2^k}\big |} = 1/2$? $$\lim_{k\to\infty} \sqrt[k] {\left | \frac{k^{1/k}-1}{2^k}\right |} = \frac12$$ according to a solution I have, but I get the limit to equal zero.
How come it goes to $\frac12$?
 A: Let
$$ a_k=\frac{k^{\frac1k}-1}{2^k}$$
thus
$$\frac{a_{k+1}}{a_k}=\frac{(k+1)^{\frac1{k+1}}-1}{2^{k+1}}\frac{2^k}{k^{\frac1k}-1}=\frac12\frac{(k+1)^{\frac1{k+1}}-1}{k^{\frac1k}-1}\to \frac12\cdot 1=\frac12$$
indeed
$$k^{\frac1k}-1=e^{\frac{\ln k}{k}}-1=\frac{\ln k}{k}+o\left(\frac{\ln k}{k}\right)$$
$${(k+1)}^{\frac1{k+1}}-1=e^{\frac{\ln (k+1)}{k+1}}-1=\frac{\ln (k+1)}{k+1}+o\left(\frac{\ln k}{k}\right)$$
$$\frac{(k+1)^{\frac1{k+1}}-1}{k^{\frac1k}-1}=\frac{\frac{\ln (k+1)}{k+1}+o\left(\frac{\ln k}{k}\right)}{\frac{\ln k}{k}+o\left(\frac{\ln k}{k}\right)}=\frac{\frac{k}{\ln k}\frac{\ln (k+1)}{k+1}+o\left(1\right)}{1+o\left(1\right)}\to 1$$
therefore
$$\frac{a_{k+1}}{a_k}\to \frac12 \implies \sqrt[k] {a_k}\to \frac12$$
A: 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,
$$
\lim_{k\to\infty}\left(\frac{k^{1/k}-1}{2^k}\right)^{1/k}=\frac12
$$
A: write $$e^{\lim_{k\to \infty}\frac{\ln(k^{1/k}-1)}{k}}$$ and use L'Hospital
A: We have
$$\sqrt[k]{\frac{\sqrt[k]{k}-1}{2^k}} = \left(\frac{\sqrt[k]{k}-1}{2^k}\right)^{1/k} = \frac{\sqrt[k]{\sqrt[k]{k} - 1}}{2}$$
Now notice that $\lim_{k\to\infty} \sqrt[k]{\sqrt[k]{k} - 1} = 1$.
Indeed, $\sqrt[k]{\sqrt[k]{k}} = k^{1/k^2} \xrightarrow{k\to\infty} 1$ because:
$$k^{1/k^2}\cdot k^{1/k^2} = (k^2)^{1/k^2} \xrightarrow{k\to\infty} 1 \implies \lim_{k\to\infty} k^{1/k^2} = 1$$
Now using the squeeze theorem we get:
$$1 \xleftarrow{k\to\infty}\sqrt[k]{\frac12}\cdot\sqrt[k]{\sqrt[k]{k}} = \sqrt[k]{\frac12\sqrt[k]{k} } \le \sqrt[k]{\sqrt[k]{k} - 1} \le \sqrt[k]{\sqrt[k]{k}} \xrightarrow{n\to\infty} 1$$
So $\lim_{k\to\infty} \sqrt[k]{\sqrt[k]{k} - 1} = 1$.
Therefore, 
$$\lim_{k\to\infty}\sqrt[k]{\frac{\sqrt[k]{k}-1}{2^k}} = \frac{\sqrt[k]{\sqrt[k]{k} - 1}}{2} = \frac12$$
A: Note that
$$k^{\frac1k}-1=e^{\frac{\ln k}{k}}-1=\frac{\ln k}{k}+o\left(\frac{\ln k}{k}\right)$$
and
$$\left(k^{\frac1k}-1\right)^{\frac1k}=\left(\frac{\ln k}{k}+o\left(\frac{\ln k}{k}\right)\right)^{\frac1k}=e^\frac{{\ln\left({\frac{\ln k}{k}+o\left(\frac{\ln k}{k}\right)}\right)}}{k}=e^{\frac{1}{\ln k}\frac{\ln k}{k}\ln\left({\frac{\ln k}{k}+o\left(\frac{\ln k}{k}\right)}\right)}\to e^0=1$$
thus
$$\sqrt[k] {\left| \frac{k^{1/k}-1}{2^k}\right|}=\frac{\sqrt[k] {\left| k^{1/k}-1\right|}}{2}\to\frac12$$
A: We can pull out a factor of $1/2.$ Note that $k^{1/k} = e^{(\ln k)/k}.$ 
Now for small positive $x,$ $x/2 < e^x - 1< 2x.$ This comes right out of the definition of $(e^x)'(0).$ Since $(\ln k)/k \to 0^+,$ this inequality implies
$$(\ln k)/2k < e^{(\ln k)/k} - 1 < 2(\ln k)/k$$
for large $k.$ Again for or large $k,$ the term on the left is greater than $1/(2k),$ and the term on the right is less than $1.$ It follows that
$$(1/(2k))^{1/k} < (k^{1/k} - 1)^{1/k} < 1^{1/k}=1$$
for large $k.$ The left side $\to 1,$ so by the squeeze theorem, the middle term $\to 1.$ Hence the desired limit is $1/2.$
