Riemann sum : sum_riemann

### 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 32n3/16n4−k4.
Compute limn → +∞ Sn.
Input :

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

Output :

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