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How can I find this limit? I've tried to use Stolz theorem, but have not succeed. I have heard smth about Riemann sums, but have not found good algorithm how to use it. Can you help me to solve it with the help of riemann sums or show me algorithm how to use it

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  • $\begingroup$ Hint: your sum is $\frac{1}{n}\sum_{k=1}^{100n}(k/n)^p$. $\endgroup$ – Wojowu Mar 13 at 16:25
  • $\begingroup$ @Wojowu Ok, I have seen this, but I am confused by 100n in sum $\endgroup$ – ErlGrey Mar 13 at 16:27
  • $\begingroup$ @ErlGrey Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Mar 25 at 17:49
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To expand on the hint in the comments: you can rewrite the sum further as $$(100)^{p+1} \frac 1{100n} \sum_{k=1}^{100n} \left( \frac k{100n} \right)^p$$ which is the Riemann sum of the function $f(x) = (100)^{p+1} x^p$ on the interval $[0,1]$ with $100n$ equal subintervals and right endpoints.

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I thought it might be instructive to present an approach that does not use Riemann sums. To that end, we proceed.


Note that

$$\sum_{k=1}^N \underbrace{\left(k^{p+1}-(k-1)^{p+1}\right)}_{=(p+1)k^p+O(k^{p-1})}=N^{p+1}$$

which by induction reveals that

$$\sum_{k=1}^N k^p=\frac{N^{p+1}}{p+1}+O(N^p)\tag1$$

Hence, we have

$$\sum_{k=1}^{100n}\frac{k^{p}}{n^{p+1}}=\frac{100^{p+1}}{p+1}+O\left(\frac1n\right)$$

Now let $n\to\infty$

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  • $\begingroup$ @ErlGrey Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit. $\endgroup$ – Mark Viola Mar 25 at 17:49
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Hint:

$$\lim_{n\to \infty}\dfrac{1}{n}\sum_{k=1}^{100n}\left(\dfrac{k}{n}\right)^p=\int_{0}^{100}x^p\mathrm dx$$


For the general case: $$\lim_{n\to \infty}\dfrac{1}{n}\sum_{k=0}^{\alpha n}f\left(\dfrac{k}{n}\right)=\int_{0}^{\alpha}f(x)\mathrm dx \ , \ \lim_{n\to \infty} \dfrac{k}{n}\Biggl|_{k=\alpha n}=\alpha $$

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