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2.18.3  Riemann sum : sum_riemann

sum_riemann takes two arguments : an expression depending of two variables and the list of the name of these two variables.
sum_riemann(expression(n,k),[n,k]) returns in the neighboorhoud 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 as a Riemann sum associated to a continue function defined on [0,1] or returns "it is probably not a Riemann sum" when the resarch is unavailing.
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 32n3/16n4k4.
Compute limn → +∞ Sn.
Input :

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

Output :

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

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