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?
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 $$ \sum_{k=2}^{n} \frac{1}{k} < \int_{1}^{n} \frac{1}{x} dx = \log n. $$
