 
 
 
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 :
^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 :
 
 
