3
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ The limits for the sum should probably be $k=1$ to $n$ since there is the $\frac1k$. $\endgroup$ – Math1000 Nov 3 '17 at 4:51
  • $\begingroup$ @Math1000 yes, fixed. Thank you. $\endgroup$ – Masacroso Nov 3 '17 at 4:53
1
$\begingroup$

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 $$\lim_{n\to\infty}\frac1{n^{m+1}}\sum_{k=1}^nk^{m+\frac1k}= \lim_{n\to\infty}\frac1{n^{\alpha}}\sum_{k=1}^n f(k/n)a_k=L\int_0^1 \alpha x^{\alpha-1}f(x)\,dx=\frac{1}{m+1}.$$

$\endgroup$
1
$\begingroup$

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 $1/2$.

$\endgroup$
0
$\begingroup$

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 $$ \lim_{n\to+\infty} \frac{1}{n^2}\sum_{k=1}^n k^{1+1/k} = \lim_{n\to+\infty} \frac{1}{n^2}\sum_{k=1}^n k = \lim_{n\to+\infty}\frac{n(n+1)}{2n^2} = \frac{1}{2}\,. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.