How to solve $\lim_{n \to \infty} \frac{n+n^{\frac{1}{2}}+n^{\frac{1}{3}}+...+n^{\frac{1}{n}}}{n}$ I have tried $\lim u^v=\lim \exp{(u-1)v}$ but i dont know how to do next..
$$\lim_{n\to \infty} \frac{1}{n}\sum_{i=1}^{n} n^{\frac{1}{i}}$$ and i have tried to find $\frac{i}{n}$but it doesn't work.
 A: Here's a quick response. Can elaborate if needed.
The thing to note is that $n^\frac{1}{i}=e^{\frac{\ln n}{i}}$ decreases fast towards $1$ as $i$ increases.
Let's first split the sum into three parts:
$$
\frac{n}{n}
 +\frac{n^{\frac{1}{2}}+\cdots+n^{\frac{1}{k}}}{n}
 +\frac{n^{\frac{1}{k+1}}+\cdots+n^{\frac{1}{n}}}{n}.
$$
The first fraction is always $1$. We now aim for the last fraction to contain the terms $n^{\frac{1}{i}}$ for which $i$ is high enough to make them all close to $1$; we then want the middle fraction to contain the terms $n^{\frac{1}{i}}$ for low $i$ which are not low enough to be included in the last fraction, but few enough of them to ensure it vanishes as $n\rightarrow\infty$.
Since the $n^{\frac{1}{i}}\le\sqrt{2}$ for all $i>1$, if $k$ increases slowly enough as $n$ increases, we can make this vanish as $n$ increases. In particular,
$$
\frac{n^{\frac{1}{2}}+\cdots+n^{\frac{1}{k}}}{n}
<\frac{k}{\sqrt{n}}
$$
so a sequence $k_n>0$ with $k_n/\sqrt{n}\rightarrow 0$ will do.
In the last fraction, all terms $n^\frac{1}{i}$ lie between $1$ and $n^\frac{1}{k}=e^\frac{\ln n}{k}$. So long as $\ln n/k_n\rightarrow 0$, the terms in the denominator will all converge to $1$, and since there are $n-k$ of them, the fraction as a whole will also converge to $1$.
So, what we need is a sequence $k_n$ so that $k_n/\sqrt{n}\rightarrow 0$ and $\ln n/k_n\rightarrow n$ as $n\rightarrow\infty$. Any $k_n\approx n^a$ for $0<a<1/2$ will do that job.
