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 $\tt 1/(n\verb\vert^\vert 3-n)$.
  Input :
  partfrac(1/(n^3-n))
  Output :
  1/(2*(n+1))-1/n+1/(2*(n-1))
  Hence :
  $$\sum_{{n=2}}^{N}$$ - $${\frac{{1}}{{n}}}$$ = - $$\sum_{{n=1}}^{{N-1}}$$$${\frac{{1}}{{n+1}}}$$ = - $${\frac{{1}}{{2}}}$$ - $$\sum_{{n=2}}^{{N-2}}$$$${\frac{{1}}{{n+1}}}$$ - $${\frac{{1}}{{N}}}$$
  $${\frac{{1}}{{2}}}$$*$$\sum_{{n=2}}^{N}$$$${\frac{{1}}{{n-1}}}$$ = $${\frac{{1}}{{2}}}$$*($$\sum_{{n=0}}^{{N-2}}$$$${\frac{{1}}{{n+1}}}$$) = $${\frac{{1}}{{2}}}$$*(1 + $${\frac{{1}}{{2}}}$$ + $$\sum_{{n=2}}^{{N-2}}$$$${\frac{{1}}{{n+1}}}$$)
  $${\frac{{1}}{{2}}}$$*$$\sum_{{n=2}}^{N}$$$${\frac{{1}}{{n+1}}}$$ = $${\frac{{1}}{{2}}}$$*($$\sum_{{n=2}}^{{N-2}}$$$${\frac{{1}}{{n+1}}}$$ + $${\frac{{1}}{{N}}}$$ + $${\frac{{1}}{{N+1}}}$$)
  After simplification by $\sum_{{n=2}}^{{N-2}}$, it remains :
  - $${\frac{{1}}{{2}}}$$ + $${\frac{{1}}{{2}}}$$*(1 + $${\frac{{1}}{{2}}}$$) - $${\frac{{1}}{{N}}}$$ + $${\frac{{1}}{{2}}}$$*($${\frac{{1}}{{N}}}$$ + $${\frac{{1}}{{N+1}}}$$) = $${\frac{{1}}{{4}}}$$ - $${\frac{{1}}{{2N(N+1)}}}$$
  Therefore :
  • for N = 10 the sum is equal to : 1/4 - 1/220 = 27/110
  • for N = + $\infty$ the sum is equal to : 1/4 because ${\frac{{1}}{{2N(N+1)}}}$ 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+1} - G_{| x=n} = f_{| x=n}.
  Input :
  sum(1/(x*(x+1)),x)
  Output :
  -1/x