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 :
^
2/n^
3,[n,k])Output :
Exercise 2
Suppose Sn=∑k=1n k3/n4.
Compute limn → +∞ Sn.
Input :
^
3/n^
4,[n,k])Output :
Exercise 3
Compute
limn → +∞(1/n+1+1/n+2+...+1/n+n).
Input :
Output :
Exercise 4
Suppose Sn=∑k=1n 32n3/16n4−k4.
Compute limn → +∞ Sn.
Input :
^
3/(16*n^
4-k^
4),[n,k])Output :