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=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.

Examples.

• Input:
  sum(1,k,-2,n)
  Output:

  n+1+2

• Input:
  normal(sum(2*k-1,k,1,n))
  Output:

  n2

• Input:
  sum(1/(n^2),n,1,10)
  Output:

  1968329 1270080

• Input:
  sum(1/(n^2),n,1,+(infinity))
  Output:

  1 6

  π 2

• Input:
  sum(1/(n^3-n),n,2,10)
  Output:

  27 110

• Input:
  sum(1/(n^3-n),n,2,+(infinity))
  Output:

  1 4

  This result comes from the decomposition of 1/(n^3-n) (see Section 6.32.9).
  Input:
  partfrac(1/(n^3-n))
  Output:

  −

  1 n

  +

  1 2  ⎛ ⎝ n−1 ⎞ ⎠

  +

  1 2  ⎛ ⎝ n+1 ⎞ ⎠

  Hence:
  ∑_n=2^N −1/n=−∑_n=1^N−1 1/n+1=−1/2−∑_n=2^N−2 1/n+1−1/N
  1/2∑_n=2^N 1/n−1=1/2(∑_n=0^N−2 1/n+1)=1/2(1+1/2+∑_n=2^N−21/n+1)
  1/2∑_n=2^N 1/n+1=1/2(∑_n=2^N−2 1/n+1+1/N+1/N+1)
  After simplification by ∑_n=2^N−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:

Output: