How to find limit of sum $\lim\limits_{n\to\infty}\sum_{k=1}^{100n}\frac{k^p}{n^{p+1}}$

Question

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

Answer 1

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.

Answer 2

Hint:

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

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

Answer 3

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

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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 $n\to\infty$