Solve limit with Riemann Sum

Question

$$\lim_{n\to \infty}\frac 1n\sum_{k=1}^n \frac{(k-1)^sk^p}{n^{s+p}}$$ I think it’s $\int_0^1x^{ s+p}dx.$ but i m not sure about.

Answer 1

hint

If $p=s=1,$ it gives

$$\frac 1n\sum_{k=1}^n\frac{k^2-k}{n^2}=$$

$$\frac 1n(\sum_{k=1}^n\frac{ k^2}{n^2}-\frac{n(n+1)}{2n^2})$$

Answer 2

Your are right, you could be sure seeing below. $$S=\lim_{n \rightarrow \infty} \frac{1}{n}\frac{(k-1)^s k^p}{n^{s+p}}$$ $$= \lim_{x \rightarrow \infty} \frac{1}{n}(k/n-1/n)^s (k/n)^p= \int_{0}^{1} x^{s+p} dx =\frac{1}{s+p+1}.$$ When $n$ is very large we use $1/n \rightarrow dx, k/n \rightarrow x$, $1/n$ in the parenthesis is ignoored. The suumation goes over to integral.

Answer 3

The best option is to squeeze the expression under limit as $$\frac{1}{n}\sum_{k=1}^{n}\frac {(k-1)^{s+p}}{n^{s+p}}\leq\frac {1}{n}\sum_{k=1}^{n}\frac{(k-1)^sk^p}{n^{s+p}}\leq\frac{1}{n}\sum_{k=1}^{n} \frac{k^{s+p}}{n^{s+p}}$$ where $s, p$ are positive. Both the leftmost and rightmost expressions in the above inequality act as Riemann sums for $f(x) =x^{s+p} $ over $[0,1]$ and therefore these tend to $\int_0^1 x^{s+P} \, dx=1/(s+p+1)$. And by Squeeze theorem the expression in the middle of the inequality also does the same.

Another approach is to note that the given sum under is actually a Riemann sum for $f(x) =x^{s+p} $ over $[0,1]$ with partition $$P=\{0,1/n,2/n,\dots,(n-1)/n,1\}$$ and tags $t_k$ given by $$f(t_k) =\frac{(k-1)^{s}k^p}{n^{s+p}}$$ (just check that $t_k\in[(k-1)/n,k/n]$).

Another slightly more difficult approach is the following theorem which says that

Let $f, g$ be real valued functions defined on $[a, b] $ and let us assume they are Riemann integrable on $[a, b] $. Then $$\int_{a} ^{b} f(x) g(x) \, dx=\lim_{||P||\to 0}\sum_{k=1}^{n}f(t_k)g(t'_k)(x_k-x_{k-1})$$ where $$P=\{x_0,x_1,x_2,\dots,x_n\}$$ is a partition of $[a, b] $ and $t_k, t'_k$ are any tag points in $[x_{k-1},x_k]$.

A proof for a more general version of the theorem is available here.

Thus for an integral of product of two functions one can choose different tags for the two functions while making a Riemann sum. Thus here the sum under limit in question uses tags $(k-1)/n$ for $f(x) =x^s$ and tags $k/n$ for $g(x) =x^p$.