6.20.3 Discrete summation: sum
The sum command can evaluate sums, series, and find discrete
antiderivatives. A discrete antiderivative of a sum ∑nf(n) is
an expression G such that G|x=n+1−G|x=n=f(n), which means
that ∑n=MNf(n) = G|x=N+1−G|M.
To evaluate a sum or series:
-
sum takes four arguments:
-
expr, an expression.
- k, the name of the variable.
- n0 and n1, integers (the bounds of the sum).
- sum(expr,k,n0,n1) returns the sum
∑k=n0n1expr.
Examples.
-
Input:
sum(1,k,-2,n)
Output:
- Input:
normal(sum(2*k-1,k,1,n))
Output:
- Input:
sum(1/(n^2),n,1,10)
Output:
- Input:
sum(1/(n^2),n,1,+(infinity))
Output:
- Input:
sum(1/(n^3-n),n,2,10)
Output:
- Input:
sum(1/(n^3-n),n,2,+(infinity))
Output:
This result comes from the decomposition of 1/(n^3-n) (see
Section 6.32.9).
Input:
partfrac(1/(n^3-n))
Output:
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.
To find a discrete antiderivative:
-
sum takes two arguments:
-
expr, an expression.
- k, the name of the variable.
- sum(expr,x) returns a discrete
antiderivative.
Example.
Input:
sum(1/(x*(x+1)),x)
Output: