### 5.18.2  Discrete summation: sum

sum takes two or four arguments :

• four arguments
an expression, the name of the variable (for example n), and the bounds (for example a and b).
sum returns the discrete sum of this expression with respect to the variable from a to b.
Input :
sum(1,k,-2,n)
Output :
n+1+2
Input :
normal(sum(2*k-1,k,1,n))
Output :
n`^`2
Input :
sum(1/(n`^`2),n,1,10)
Output :
1968329/1270080
Input :
sum(1/(n`^`2),n,1,+(infinity))
Output :
pi`^`2/6
Input :
sum(1/(n`^`3-n),n,2,10)
Output :
27/110
Input :
sum(1/(n`^`3-n),n,1,+(infinity))
Output :
1/4
This result comes from the decomposition of 1/(n`^`3-n).
Input :
partfrac(1/(n`^`3-n))
Output :
1/(2*(n+1))-1/n+1/(2*(n-1))
Hence :
n=2N −1/n=−∑n=1N−1 1/n+1=−1/2−∑n=2N−2 1/n+1−1/N
1/2*∑n=2N 1/n−1=1/2*(∑n=0N−2 1/n+1)=1/2*(1+1/2+∑n=2N−21/n+1)
1/2*∑n=2N 1/n+1=1/2*(∑n=2N−2 1/n+1+1/N+1/N+1)
After simplification by ∑n=2N−2, it remains :
−1/2+1/2*(1+1/2)−1/N+1/2*(1/N+1/N+1)=1/4−1/2N(N+1)
Therefore :
• for N=10 the sum is equal to : 1/4−1/220=27/110
• for N=+∞ the sum is equal to : 1/4 because 1/2N(N+1) approaches zero when N approaches infinity.
• two arguments
an expression of one variable (for example f) and the name of this variable (for example x).
sum returns the discrete antiderivative of this expression, i.e. an expression G such that G|x=n+1G|x=n=f|x=n.
Input :
sum(1/(x*(x+1)),x)
Output :
-1/x