Calculating $\lim_{n\to\infty}\sqrt[n]{ \sqrt[n]{n} - 1 }$ I know that   
$$\lim_{n\to\infty} \sqrt[n]{ \sqrt[n]{n} - 1 } = 1,$$
but I'm unable to prove it. I could easily estimate that it's at most $1$, but my best estimation from below is that the limit is greater than $0$.
Doing this from the definition doesn't lead me anywhere either.
 A: Since deriving stuff like $f(x)^{g(x)}$ is a nightmare (at least to me), always try to apply the logarithm in these cases, and see if you can come up with something easier.
$$\lim_{n \to \infty} (n^{\frac{1}{n}} -1)^{\frac{1}{n}}$$
$$\lim_{n \to \infty} \log(n^{\frac{1}{n}} -1)^{\frac{1}{n}} = \lim_{n\to \infty} \frac{\log (n^{\frac{1}{n}} -1)}{n}$$
$$ = \lim_{n\to \infty} \frac{\log (n^{\frac{1}{n}} -1)}{n}$$
Now since
$\log (n^{\frac{1}{n}} -1) \leq (n^{\frac{1}{n}} -1) \leq n^{\frac{1}{n}}$
we get
$$\lim_{n\to \infty} \frac{\log (n^{\frac{1}{n}} -1)}{n} \leq \lim_{n \to \infty} \frac{1}{n^{\frac{n-1}{n}}} = 0$$
Which means that the above limit is $1$, since $\log(1)=0$.
A: The term being limited is
$$\left (e^{\log{n}/n}-1 \right )^{1/n} =  \left (\frac{\log{n}}{n} + \cdots\right )^{1/n}$$
So consider
$$\lim_{n \to \infty}\frac1n \log{\left (\frac{\log{n}}{n} \right )} = \lim_{n \to \infty} \frac{n}{\log{n}} \frac{1-\log{n}}{n^2} = 0$$
by L'Hopital.  This is the log of the limit.  The limit we seek is therefore $1$.
A: Using $ \displaystyle 1+\frac1n \lt \sqrt[n]{n} \lt 1+\sqrt{\frac1n} $ for $n \ge 3$ and thus $ \displaystyle \frac{1}{ 1+\sqrt{\frac1n}} \lt \sqrt[n]{\frac1n} \lt \frac{1}{ 1+{\frac1n}}$ 
gives $ \displaystyle  \frac1n \lt \sqrt[n]{n}-1 \lt \sqrt{\frac1n} $
so $ \displaystyle \frac{1}{ 1+\sqrt{\frac1n}} \lt \sqrt[n]{{\frac1n}} \lt \sqrt[n]{\sqrt[n]{n}-1} \lt \sqrt[n]{\sqrt{\frac1n}} \lt \sqrt{\frac{1}{ 1+{\frac1n}}}$ 
with the left- and right-hand expressions each converging towards $1 $ as $n$ increases 
A: The numbers $n^{k/n}$ with $0\leq k\leq n-1$ are all between $1$ and $n$. By the formula for the sum of finite  geometric series it follows that
$$n\leq\sum_{k=0}^{n-1} n^{k/n}={n-1\over n^{1/n}-1}\leq n^2\qquad(n\geq2)\ .$$
From this we infer
$${1\over 2n}<{n-1\over n^2}\leq n^{1/n}-1\leq{n-1\over n}<1\ .$$
Using   $\lim_{n\to\infty} (2n)^{1/n}=1$ and the squeeze theorem one then concludes that $$\lim_{n\to\infty}\bigl(n^{1/n}-1)^{1/n}=1\ .$$ 
A: Idea: extend to $\Bbb R$, take $\log$:
$$\lim_{x\to+\infty}\frac{\log(x^{1/x}-1)}x$$
and use L'Hôpital.
A: Only Taylor series, only hardcore)
1) $n^{\frac{1}{n}} = e^{\frac{\log n}{n}} \sim 1 + \frac{\log n}{n}$, so the expression under the first root is $\frac{\log n }{n}$
2)$\big(\frac{\log n}{n} \big)^\frac{1}{n} = e^{\frac{\log \frac{\log n}{n}}{n}}  = e^{\frac{\log \log n}{n}} \cdot e^{-\frac{\log n }{n}} \sim (1+\frac{\log \log n}{n})(1-\frac{\log n }{n}) \to_n 1$
A: We can manipulate limits as long as we are sure that the part we are jumping over is bounded. For example:
$$\lim\limits_{x \to \infty} f(x)^{g(x)} = 
\lim\limits_{x \to \infty} f(x)^{\lim\limits_{x \to \infty}g(x)}$$
as long as $f(x)$ is bounded.
First we attack $\lim\limits_{n \to \infty} n^{\frac{1}{n}}$. We make a substitution $x=\frac{1}{n}$
$$\lim\limits_{n \to \infty} n^{\frac{1}{n}}=\lim\limits_{x \to 0} (\frac{1}{x})^x=\lim\limits_{x \to 0} \frac{1}{x^x}$$
Now this one is elementary $\lim\limits_{x \to 0} x^x = 1$ so we have $\lim\limits_{n \to \infty} n^{\frac{1}{n}}=1$
We continue
$$\lim\limits_{n \to \infty} \sqrt[n]{\sqrt[n]{n}-1}=\lim\limits_{n \to \infty} \sqrt[n]{\sqrt[n]{n}(1-\frac{1}{\sqrt[n]{n}})}=\lim\limits_{n \to \infty}n^{\frac{1}{n^2}} (1-\frac{1}{n^{\frac{1}{n}}})^{\frac{1}{n}}$$
Since $n^{\frac{1}{n^2}}=(n^{\frac{1}{n}})^{\frac{1}{n}}$ we have as well $\lim\limits_{n \to \infty} n^{\frac{1}{n^2}}=1$
That leaves
$$\lim\limits_{n \to \infty} \sqrt[n]{\sqrt[n]{n}-1}=\lim\limits_{n \to \infty} (1-(\frac{1}{n})^{\frac{1}{n}})^{\frac{1}{n}}$$
The same substitution and we have 
$$\lim\limits_{x \to 0} (1-x^x)^{x}$$
$x^x$ has a minimum at $\frac{1}{e}$ and $1 > 1-x^x > 0$ for $0 < x < 1$ meaning it is bounded and $1-x^x$ is not reaching 0 for any $x$ in this region so we can examine
$$\lim\limits_{x \to 0} (1-x^x)^{x}=\lim\limits_{x \to 0} x^{x^2}(\frac{1}{x^x}-1)^{x}=\lim\limits_{x \to 0} (\frac{1}{x^x}-1)^{x}$$
This is because we can write $x^{x^2}=(x^x)^x$
Obviously both limits are equal $\lim\limits_{x \to 0} (\frac{1}{x^x}-1)^{x}=\lim\limits_{x \to 0} (1-x^x)^{x}$.
Notice that
$$\lim\limits_{x \to 0} (\frac{1}{x^x}-1)^{x}=\lim\limits_{x \to 0} (x^{-x}-1)^{x}=\lim\limits_{x \to 0^{-}} (x^{x}-1)^{-x}=\lim\limits_{x \to 0^{-}} \frac{1}{(x^{x}-1)^{x}}$$
which is to say that the right limit is equal to reciprocal left (if it exists). This means that the unique limit, if it exists, is 1 since left and right limits are then the same.
But, $(1-x^x)^x$ is continuous at 0, $x^x$ is continuous and $a^x$ is continuous, meaning the limit is indeed 1.
A: $$\lim _{n\to \infty \:}\left(\left(n^{\frac{1}{n}}-1\right)^{\frac{1}{n}}\right) = \lim _{n\to \infty \:}\left(e^{\frac{1}{n}\ln \left(n^{\frac{1}{n}}-1\right)}\right)$$
Apply the Limit Chain Rule
$$\mathrm{if}\:\lim _{u\:\to \:b}\:f\left(u\right)=L,\:\mathrm{and}\:\lim _{x\:\to \:a}g\left(x\right)=b,\:\mathrm{and}\:f\left(x\right)\:\mathrm{is\:continuous\:at}\:x=b$$
$$\mathrm{Then:}\:\lim _{x\:\to \:a}\:f\left(g\left(x\right)\right)=L$$
$$g\left(n\right)=\frac{1}{n}\ln \left(n^{\frac{1}{n}}-1\right),\:f\left(u\right)=e^u$$
So
$$\lim _{n\to \infty \:}\left(\frac{1}{n}\ln \left(n^{\frac{1}{n}}-1\right)\right)=\lim _{n\to \infty \:}\left(\frac{\ln \left(n^{\frac{1}{n}}-1\right)}{n}\right)$$
$\mathrm{Apply\:L'Hopital's\:Rule}\rightarrow\lim _{n\to \infty \:}\left(\frac{\frac{1-\ln \left(n\right)}{n^2-n^{2-\frac{1}{n}}}}{1}\right)=\lim _{n\to \infty \:}\left(\frac{-n^{\frac{1}{n}-2}\left(\ln \left(n\right)-1\right)}{n^{\frac{1}{n}}-1}\right)$
$\mathrm{Apply\:L'Hopital's\:Rule}\rightarrow\lim _{n\to \infty \:}\left(\frac{n^{\frac{1}{n}-4}\left(\left(\ln \left(n\right)-1\right)^2+n\left(2\ln \left(n\right)-3\right)\right)}{n^{\frac{1}{n}-2}\left(-\left(\ln \left(n\right)-1\right)\right)}\right)= \lim _{n\to \infty \:}\left(\frac{3n-\ln ^2\left(n\right)-2n\ln \left(n\right)+2\ln \left(n\right)-1}{n^2\left(\ln \left(n\right)-1\right)}\right)$
$\mathrm{Apply\:L'Hopital's\:Rule}\rightarrow\lim _{n\to \infty \:}\left(\frac{\frac{2-2\ln \left(n\right)}{n}-2\ln \left(n\right)+1}{n\left(2\ln \left(n\right)-1\right)}\right)=\lim _{n\to \infty \:}\left(\frac{n-2n\ln \left(n\right)-2\ln \left(n\right)+2}{n^2\left(2\ln \left(n\right)-1\right)}\right)$
$\mathrm{Apply\:L'Hopital's\:Rule}\rightarrow\lim _{n\to \infty \:}\left(\frac{-\frac{2}{n}-2\ln \left(n\right)-1}{4n\ln \left(n\right)}\right)=\lim _{n\to \infty \:}\left(\frac{-n-2n\ln \left(n\right)-2}{4n^2\ln \left(n\right)}\right)$
$\mathrm{Apply\:L'Hopital's\:Rule}\rightarrow\lim _{n\to \infty \:}\left(\frac{-2\ln \left(n\right)-3}{4n\left(2\ln \left(n\right)+1\right)}\right)$
$\mathrm{Apply\:L'Hopital's\:Rule}\rightarrow\lim _{n\to \infty \:}\left(\frac{-\frac{2}{n}}{8\ln \left(n\right)+12}\right)$
Then
$$=\lim _{n\to \infty \:}\left(\frac{-1}{2n\left(2\ln \left(n\right)+3\right)}\right)$$
Now
$$\lim _{x\to a}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\lim _{x\to a}f\left(x\right)}{\lim _{x\to a}g\left(x\right)}\mathrm{,\:where\:}\lim _{x\to a}g\left(x\right)\ne 0$$
$$\frac{\lim _{n\to \infty \:}\left(-1\right)}{\lim _{n\to \infty \:}\left(2n\left(2\ln \left(n\right)+3\right)\right)}=\frac{-1}{\infty \:}=0$$
So 
$$\lim _{u\to \:0}\left(e^u\right)=1$$
Finally
$$\lim _{n\to \infty \:}\left(\left(n^{\frac{1}{n}}-1\right)^{\frac{1}{n}}\right)=1$$
