Calculating $\lim_{n\to\infty}\sqrt[n]{ \sqrt[n]{n} - 1 }$

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 $$\lim_{n\to\infty}\bigl(n^{1/n}-1)^{1/n}=1\ .$$

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 $\log(1)=0$.

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 $1$.