evaluate $\lim_{n\rightarrow\infty}\frac{1}{n}\sum^n_{k=1} k^{\frac{1}{k}}$

Question

evaluate $\lim_{n\rightarrow\infty}\frac{1}{n}\sum^n_{k=1} k^{\frac{1}{k}}$

I could guess that it converges to 1 but I'm having trouble showing how

Answer 1

As already noted, we could appeal to the Stolz-Cesaro theorem. Instead, we go back to the basics to evaluate the limit.

First we note that $\lim_{k\to \infty}k^{1/k}=1$. Therefore, given any $\epsilon >0$, there exists a number $N$ such that whenever $k>N$, we have

$$1-\epsilon <k^{1/k}<1+\epsilon \tag 1$$

With that value of $\epsilon$ and $N$, we write the sum of interest as

$$\frac1n \sum_{k=1}^nk^{1/k}=\frac 1n\sum_{k=1}^Nk^{1/k}+\frac 1n \sum_{N+1}^nk^{1/k} \tag 2$$

For the fixed $N$, the first sum on the right-hand of $(2)$ goes to zero as $n\to \infty$. For the second sum, we have using $(1)$

$$\left(1-\frac{N+1}{n}\right)(1-\epsilon)<\frac 1n \sum_{N+1}^nk^{1/k}<\left(1+\frac{N+1}{n}\right)(1+\epsilon) \tag 3$$

From $(3)$ it is clear that we can choose $n$ so large that the sum $\frac 1n \sum_{N+1}^nk^{1/k}$ is arbitrarily close to $1$. And we are done!

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(\frac1n \sum_{k=1}^nk^{1/k}\right)=1}$$

Answer 2

Hint: Try to use (or prove) the following theorem.

If $\lim_{n\to\infty}a_n=a$, then $\lim_{n\to\infty}\frac1n(a_1+\cdots+a_n)=a$.

Answer 3

Another way is to note that, by Abel's summation, $$\sum_{k\leq n}k^{1/k}=\sum_{k\leq n}1\cdot k^{1/k}=n\cdot n^{1/n}+\int_{1}^{n}\left\lfloor t\right\rfloor t^{1/t-2}\left(\log\left(t\right)-1\right)dt$$ and so, using $\left\lfloor t\right\rfloor =t+O\left(1\right)$ we have $$ =n\cdot n^{1/n}+\int_{1}^{n}t^{1/t-1}\left(\log\left(t\right)-1\right)dt+O\left(n^{1/n}\right)=$$ $$=n\cdot n^{1/n}+\int_{1}^{n}\frac{t^{1/t}\log\left(t\right)dt}{t}-\int_{1}^{n}\frac{t^{1/t}dt}{t}+O\left(n^{1/n}\right)$$ now using $1\leq t^{1/t}\leq e^{1/e}$ we can estimate the integrals $$\frac{\log^{2}\left(n\right)}{2}\leq\int_{1}^{n}\frac{t^{1/t}\log\left(t\right)dt}{t}\leq e^{1/e}\frac{\log^{2}\left(n\right)}{2}$$ and $$\log\left(n\right)\leq\int_{1}^{n}\frac{t^{1/t}dt}{t}\leq e^{1/e}\log\left(n\right)$$ hence $$n^{\frac{n+1}{n}}+\frac{\log^{2}\left(n\right)}{2}-e^{1/e}\log\left(n\right)+O\left(n^{1/n}\right)\leq\sum_{k\leq n}k^{1/k}\leq n^{\frac{n+1}{n}}+e^{1/e}\frac{\log^{2}\left(n\right)}{2}-\log\left(n\right)+O\left(n^{1/n}\right)$$ and so $$\frac{1}{n}\sum_{k\leq n}k^{1/k}\underset{n\rightarrow\infty}{\longrightarrow}1.$$