Find the limit of sequence $a_{n+1} -a_{n}$ 
Consider the sequence $a_{n}=\sqrt{n}+\sqrt[3]{n}+\cdots+ \sqrt[n]{n}.$ Find the limit of sequence $a_{n+1} -a_{n}.$

One of my attempts is to use the reciprocal of the Stolz-Cesaro lemma, which is true by imposing additional conditions. What is the valid solution?
 A: We will show that the limit is 1. (I actually do some experiment with programming and find that it looks like converges to 1.)
We have
$$
a_{n+1} - a_{n} = ((n+1)^{1/2} - n^{1/2}) + \cdots + ((n+1)^{1/n} - n^{1/n}) + (n+1)^{1/(n+1)}
$$
and it is well-known (maybe not) that $\lim_{n\to \infty} (n+1)^{1/(n+1)} = 1$. This can be proved by taking $\log$ and applying L'Hospital's lemma. Anyway, we will prove that 
$$
\lim_{n\to \infty} \sum_{k=2}^{n}((n+1)^{1/k} - n^{1/k}) = 0.
$$
Since the sum is positive, it is enough to show that the sequence is bounded above by another sequence which converges to 0. By Mean Value Theorem, for $f(x) = x^{1/k}$, we have
$$
(n+1)^{1/k} - n^{1/k} = \frac{f(n+1) - f(n)}{(n+1) - n} = f'(c) \leq f'(n) = \frac{1}{kn^{1-1/k}} \leq \frac{1}{kn^{1/2}}
$$
for some $n <c < n+1$. Here we use that $f'(x)$ is a decreasing function for $k\geq 2$. From this, we have
$$
\sum_{k=2}^{n} ((n+1)^{1/k} - n^{1/k}) \leq \sum_{k=2}^{n} \frac{1}{kn^{1/2}} < \frac{\log n }{n^{1/2}} \xrightarrow{n\to \infty}0
$$
where $\sum_{k=2}^{\infty} 1/k < \log n$ is obtained from 
$$
\sum_{k=2}^{n} \frac{1}{k} < \int_{1}^{n} \frac{1}{x} dx = \log n.
$$
