Trying to transform a Riemann-sum-like expression into an integral

Question

This question comes from curiosity. I know that, in general:

$$\lim_{n\to\infty}\frac1n\sum_{k=0}^n f(k/n)=\int_0^1 f(x)\,\mathrm dx\tag1$$

Then I have this expression

$$\lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^nk^{1+\frac1k}\tag2$$

I know, by other methods, that the value of $(2)$ is $1/2$, but Im interested to know if its possible (or someone know how) to transform this Riemann-sum-like in an integral of Riemann for some appropriated $f$.

Indeed Im interested too to know some reference about this topic (if it exists).

Answer 1

You may take a look at Theorem 1 given in A Generalization of Riemann Sums by Omran Kouba. In your case, take $a_k=k^{1/k}$, $f(x)=x$, $\alpha=1$, and $L=1$ (see Showing $ \sum_{k=1}^{n} k^{1/k}\sim n$), then $$\lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^nk^{1+\frac1k}= \lim_{n\to\infty}\frac1{n^{\alpha}}\sum_{k=1}^n f(k/n)a_k=L\int_0^1 \alpha x^{\alpha-1}f(x)\,dx=\int_0^1 xdx=\frac{1}{2}.$$ More generally, by taking $f(x)=x^m$ we have that $$\lim_{n\to\infty}\frac1{n^{m+1}}\sum_{k=1}^nk^{m+\frac1k}= \lim_{n\to\infty}\frac1{n^{\alpha}}\sum_{k=1}^n f(k/n)a_k=L\int_0^1 \alpha x^{\alpha-1}f(x)\,dx=\frac{1}{m+1}.$$

Answer 2

I propose a direct proof.

Since $k^{1/k} > 1$ and $\lim_k k^{1/k} = 1$, given $\varepsilon > 0$ there exists $N\in\mathbb{N}$ such that $1 < k^{1/k} < 1+\varepsilon$ for every $n>N$.

Hence, for every $n>N$, we have that $$ \begin{split} 0 & \leq \frac{1}{n^2}\sum_{k=1}^n k^{1+1/k} - \frac{1}{n^2}\sum_{k=1}^n k = \frac{1}{n^2}\sum_{k=1}^N (k^{1+1/k} - k) + \frac{1}{n^2}\sum_{k=N+1}^n k(k^{1/k} - 1) \\ & \leq \frac{1}{n^2}\sum_{k=1}^N (k^{1+1/k} - k) + \frac{\varepsilon}{n^2}\sum_{k=N+1}^n k \leq \frac{1}{n^2}\sum_{k=1}^N (k^{1+1/k} - k) + \frac{\varepsilon}{2}\,, \end{split} $$ so that $$ 0\leq \limsup_{n\to +\infty} \left(\frac{1}{n^2}\sum_{k=1}^n k^{1+1/k} - \frac{1}{n^2}\sum_{k=1}^n k\right) \leq \frac{\varepsilon}{2}\,. $$ It follows that $$ \lim_{n\to+\infty} \frac{1}{n^2}\sum_{k=1}^n k^{1+1/k} = \lim_{n\to+\infty} \frac{1}{n^2}\sum_{k=1}^n k = \lim_{n\to+\infty}\frac{n(n+1)}{2n^2} = \frac{1}{2}\,. $$

Answer 3

Yet another proof:

By the Cesaro-Stolz theorem, we are led to consider the limit $$ \lim_n \frac{n^{1+1/n}}{n^2-(n-1)^2} = \lim_n \frac{n}{2n-1}\, n^{1/n} = \frac{1}{2}. $$ Since this limit exists, we can conclude that also the original limit exists and it is $1/2$.