Error in approximating the sum I am watching one of the online probability courses and in one of the lectures, the professor simplifies the sum:
$$A = \sum_{j=0}^{N}\frac{j^k}{N^k} \cdot \frac{1}{N+1}$$
in the following way: $A \approx  \int_{0}^{1}x^kdx=\frac{1}{k+1}$
The problem is that in my opinion the simplification should work in the following way:
$$A =  \frac{1}{N+1} \cdot  \sum_{j=0}^{N}\frac{j^k}{N^k} $$ and then if one will make a substitution $x = \frac{j}{N}$, then $$A \approx \frac{1}{N+1} \cdot  \int_{0}^{1}x^kdx$$ and doing completely the same I will end up with $$A = \frac{1}{N+1} \cdot \frac{1}{k+1}$$
I can not figure out where is my problem here.
 A: Using the suggestions given so far note that the role of $\Delta x_i$ is played by $\frac{1}{N+1}$.   
So when you write
$$A = \sum_{j=0}^{N}\frac{j^k}{N^k} \cdot \frac{1}{N+1}$$ keep the term involving $N+1$ inside the summation instead of pulling it outside as you did, and let $ \frac{j^k}{N^k} $ play the role of $f(w_i)$.  Then perhaps you'll see how to get to the approximation given by your professor.  And since we're only going for an approximation here, it will be close enough if $N$ is large enough.     
A: SalmonKiller gave a good point and I shall not repeat. However, one could notice that $$A_{k,N} = \sum_{j=1}^{N}\frac{j^k}{N^k} \cdot \frac{1}{N+1}=\frac{ H_N^{(-k)}}{(N+1)N^k}$$ where appear the harmonic numbers. Expanding as series for large values of $N$, we then have $$A_{k,N}=\left(\frac{1}{k+1}+\frac{\frac{1}{2}-\frac{1}{k+1}}{N}+O\left(\left(\frac{1}{N}\right)^2\right)\right)+ \frac{\zeta (-k)}{N^k} \left(\frac{1}{N}-\frac{1}{N^2}+O\left(\left(\frac{1}{N}\right)^3\right)\right)$$ So, if $k>1$, $$A_{k,N} \approx\frac{1}{k+1}+\frac{\frac{1}{2}-\frac{1}{k+1}}{N}=\frac{1}{k+1}\Big(1-\frac 1N\Big)+\frac 1{2N}$$
Let us try with small numbers $A_{5,10}=\frac{803}{4000}$ while the approximation gives $\frac{800}{4000}$.
A: Well, the issue is that $\sum_{j=0}^{N}f(x) \neq \int_{0}^{1}f(x) dx$. A Riemann sum is defined as $$\sum_{i=1}^{\infty}f(w_i)\Delta x_i$$ In your case $N \neq \infty$. So there is no way this could be a Riemann Sum construction of the definite integral.
