### 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=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.
- 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))

1
6

π2

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

27
110

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

1
4

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

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

−

1
n

+

1

2 | ⎛ ⎝ | n−1 | ⎞ ⎠

+

1

2 | ⎛ ⎝ | n+1 | ⎞ ⎠

Hence:

N
∑
n=2

−

1
n

=−

N−1
∑
n=1

1
n+1

=−

1
2

−

N−2
∑
n=2

1
n+1

−

1
N

1
2

N
∑
n=2

1
n−1

=

1
2

⎛
⎜
⎜
⎝

N−2
∑
n=0

1
n+1

)=

1
2

(1+

1
2

+

N−2
∑
n=2

1
n+1

⎞
⎟
⎟
⎠

1
2

N
∑
n=2

1
n+1

=

1
2

⎛
⎜
⎜
⎝

N−2
∑
n=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.

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

−

1
x