Discrete summation

13.3.3 Discrete summation

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=M^Nf(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=n₀^n₁expr.
To find a discrete antiderivative, sum takes two arguments:
- expr, an expression.
- x, the name of the variable.
sum(expr,x) returns a discrete antiderivative.

Examples

sum(1,k,-2,n)

n+1+2

normal(sum(2*k-1,k,1,n))

n²

sum(1/(n^2),n,1,10)

1968329

1270080

sum(1/(n^2),n,1,+(infinity))

π ²

sum(1/(n^3-n),n,2,10)

110

sum(1/(n^3-n),n,2,+(infinity))

This result comes from the decomposition of 1/(n^3-n) (see Section 11.6.9).

partfrac(1/(n^3-n))

−

⎛
⎝

n−1

⎞
⎠

⎛
⎝

n+1

⎞
⎠

Hence:

∑

n=2

−

=−

N−1

∑

n=1

n+1

=−

−

N−2

∑

n=2

n+1

−

∑

n=2

n−1

⎛
⎜
⎜
⎝

N−2

∑

n=0

n+1

(1+

N−2

∑

n=2

n+1

⎞
⎟
⎟
⎠

∑

n=2

n+1

⎛
⎜
⎜
⎝

N−2

∑

n=2

n+1

N+1

⎞
⎟
⎟
⎠

After simplification by ∑_n=2^N−2, it remains:

−

⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠

−

⎛
⎜
⎜
⎝

N+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.

sum(1/(x*(x+1)),x)

−