Limit of $\frac1n\left(1+\frac1{\sqrt[n]{2}}+\frac1{\sqrt[n]{3}}+\dotsb+\frac1{\sqrt[n]{n}}\right)$ when $n\to\infty$

Question

Calculate this limit $$\lim\limits_{n \to \infty} \frac{1}{n}\left(1+\frac{1}{\sqrt[n]{2}}+\frac{1}{\sqrt[n]{3}}+\dotsb+\frac{1}{\sqrt[n]{n}}\right).$$

I think inside the parentheses, each limit is $1$, and there are $n$ of them, so their sum is limited to $n$. Also,

$$\lim\limits_{n \to \infty}\frac{1}{n}=0.$$

Therefore I think, $$\lim\limits_{n \to \infty} \frac{1}{n}\left(1+\frac{1}{\sqrt[n]{2}}+\frac{1}{\sqrt[n]{3}}+\dotsb+\frac{1}{\sqrt[n]{n}}\right) = 0.$$

Is this solution correct? If so, how to prove it?

Answer 1

Actually the limit is $1$ because $\lim\limits_{n\to\infty}\sqrt[n]{n}=1$ and $$ \frac{n}{\sqrt[n]{n}}\leqslant1+\frac{1}{\sqrt[n]{2}}+\frac{1}{\sqrt[n]{3}}+\dotsb+\frac{1}{\sqrt[n]{n}}\leqslant n. $$ Note that your approach would also yield the limit $0$ for the sequence $$ \frac1n\cdot\left(1+1+\cdots+1\right), $$ for every number of terms in the parenthesis.

Answer 2

Let $x_n = \dfrac{1}{n^{1/n}}$, then you can show : $x_n \to 1$ by various means, then apply Cesaro theorem to get the limit $1$.

Answer 3

Using the Cezàro mean the desired limit is equal to $$\lim_{n\to\infty}\frac{1}{\sqrt[n]{n}}=1$$