i was looking at a problem my professor gave me:
$$\lim_{n\to\infty}\frac{\sum_{k=1}^n \sqrt[k]k}{n} $$ the approach i took is i looked $\sqrt[k]k$ and studying the derivative we know it is decreasing for k>e and we also know it approaches 1 for k->$\infty$. this tells me that since $\sum_{k=1}^n 1 =n$ and the numerator is always greater than the denominator my limit has to be >=1.
then i decomposed $\sum_{k=1}^n \sqrt[k]k=\sum_{k=1}^n 1+S_k$ where $S_k=\sqrt[k]k -1$
so now: $$\lim_{n\to\infty}\frac{\sum_{k=1}^n \sqrt[k]k}{n} = \lim_{n\to\infty} 1+\frac{\sum_{k=1}^n S_k }{n}$$ and if $\sum_{k=1}^n S_k$ converges then the whole term is 0 so the whole limit is equal to 1, however the series does not converge so the limit cant be 1. my intuition tells me that the limit just diverges to infinity but how do i go about proving it rigorously?
also, this makes me question, since the numerator is only slightly bigger than the denominator at every n and is actually approaching the denominator value, yet it still looks like the limit diverges how much "bigger" does a term need o be to be considered a greater infinity (e.g. $x^2\gt x$ $x->\infty$)? is there a clear line?,
Note that for $k\ge 3$, the term $\sqrt[k]{k}$ is strictly decreasing and tending to $1$. You can prove it by both calculating its derivative or by proving $(1+1/k)^k<3$ for $k\ge 3$. This means that for any $1<\alpha<\sqrt[3]3$, there is some $k(\alpha)$ for which, $k\ge k(\alpha)$ yields $\sqrt[k]k\le \alpha$.
It is obvious that the limit should be $\ge 1$, since all the terms of $\sqrt[k] k$ are $\ge 1$. Now, fix some $\alpha$ with $1<\alpha<\sqrt[3]3$. Therefore $$ \lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^n \sqrt[k]k {= \lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^{k(\alpha)-1} \sqrt[k]k+\frac{1}{n}\sum_{k=k(\alpha)}^n \sqrt[k]k \\\le 0+\lim_{n\to \infty}\frac{1}{n}\sum_{k=k(\alpha)}^n \alpha \\= 0+\lim_{n\to \infty}\frac{1}{n}(n-k(\alpha)) \alpha \\= \alpha. } $$ Hence, the limit is between $1$ and $\alpha$ for any $1<\alpha<\sqrt[3]3$. Therefore, the limit is $1 \blacksquare$.