### 5.18.3 Riemann sum : sum_riemann

sum_riemann takes two arguments : an expression depending on two variables and the list of the name of these two variables.
sum_riemann(expression(n,k),[n,k]) returns in the neighborhood of n=+∞ an equivalent of ∑k=1n expression(n,k) (or of ∑k=0n−1 expression(n,k) or of ∑k=1n−1 expression(n,k)) when the sum is looked on as a Riemann sum associated to a continuous function defined on [0,1] or returns "it is probably not a Riemann sum" when the no result is found.
Exercise 1
Suppose Sn=∑k=1n k2/n3.
Compute limn → +∞ Sn.
Input :

sum_riemann(k`^`2/n`^`3,[n,k])

Output :

1/3

Exercise 2
Suppose Sn=∑k=1n k3/n4.
Compute limn → +∞ Sn.
Input :

sum_riemann(k`^`3/n`^`4,[n,k])

Output :

1/4

Exercise 3
Compute limn → +∞(1/n+1+1/n+2+...+1/n+n).
Input :

sum_riemann(1/(n+k),[n,k])

Output :

log(2)

Exercise 4
Suppose Sn=∑k=1n 32n3/16n4−k4.
Compute limn → +∞ Sn.
Input :

sum_riemann(32*n`^`3/(16*n`^`4-k`^`4),[n,k])

Output :

2*atan(1/2)+log(3)