If the option parametric is specified, then Definite returns a result that is valid for all possible integer values of any parameters occurring in the summand or the summation bounds. In general, the result is expressed in terms of piecewise functions.

For a description of the formal option, see sum/details.

The Definite(f, k=m..n) command computes a closed form of the definite sum of $f$ over the specified range of $k$.

The function is a combination of different algorithms.  They include

the method of integral representation,

the method of first computing a closed form of the corresponding indefinite sum and then applying the discrete Newton-Leibniz formula,

the method of computing closed forms of definite sums of hypergeometric terms (see SumTools[Hypergeometric]), and

the method of first converting the given definite sum to hypergeometric functions, and then converting these hypergeometric functions to standard functions (if possible).

For more information, see sum.

$\mathrm{with}\left(\mathrm{SumTools}\left[\mathrm{DefiniteSum}\right]\right)\:$

$F≔\frac{{\left(2+k\right)}^{k-2}{\left(1+n-k\right)}^{n-k}}{k!\left(n-k\right)!}$

$F≔\frac{{\left(2+k\right)}^{k-2}{\left(1+n-k\right)}^{n-k}}{k!\left(n-k\right)!}$

$\mathrm{Sum}\left(F\,k=0..n\right)=\mathrm{Definite}\left(F\,k=0..n\right)$

$\sum _{k=0}^n\frac{{\left(2+k\right)}^{k-2}{\left(1+n-k\right)}^{n-k}}{k!\left(n-k\right)!}=\frac{{\left(3+n\right)}^n}{4n!}-\frac{{\left(3+n\right)}^{n-1}}{6\left(n-1\right)!}$

$F≔\mathrm{binomial}\left(2n-2k\,n-k\right)2^{4k}{\left(2k\left(2k+1\right)\mathrm{binomial}\left(2k\,k\right)\right)}^{-1}$

$F≔\frac{\left(\genfrac{}{}{0ex}{}{2n-2k}{n-k}\right)2^{4k}}{2k\left(2k+1\right)\left(\genfrac{}{}{0ex}{}{2k}{k}\right)}$

$\mathrm{Sum}\left(F\,k=1..n\right)=\mathrm{Definite}\left(F\,k=1..n\right)$

$\sum _{k=1}^n\frac{\left(\genfrac{}{}{0ex}{}{2n-2k}{n-k}\right)2^{4k}}{2k\left(2k+1\right)\left(\genfrac{}{}{0ex}{}{2k}{k}\right)}=-\frac{\left(\genfrac{}{}{0ex}{}{2n-2}{n-1}\right)\left(8-16n\right)}{2\left(2n+1\right)}$

$F≔{\mathrm{binomial}\left(2n\,2k\right)}^2$

$F≔{\left(\genfrac{}{}{0ex}{}{2n}{2k}\right)}^2$

$\mathrm{Sum}\left(F\,k=0..n\right)=\mathrm{Definite}\left(F\,k=0..n\right)$

$\sum _{k=0}^n{\left(\genfrac{}{}{0ex}{}{2n}{2k}\right)}^2=\frac{{\left(\mathrm{-1}\right)}^n\left(\genfrac{}{}{0ex}{}{2n}{n}\right)}{2}+\frac{\left(\genfrac{}{}{0ex}{}{4n}{2n}\right)}{2}$

$F≔\frac{\frac{2^{2k}}{{\mathrm{\pi }}^{\frac{1}{2}}}\mathrm{\Gamma }\left(k-n\right)\mathrm{\Gamma }\left(k+n\right)}{\mathrm{\Gamma }\left(2k+1\right)}{z}^k$

$F≔\frac{2^{2k}\mathrm{\Gamma }\left(-n+k\right)\mathrm{\Gamma }\left(k+n\right)z^k}{\sqrt{\mathrm{\pi }}\mathrm{\Gamma }\left(2k+1\right)}$

$\mathrm{Sum}\left(F\,k=0..\mathrm{\infty }\right)=\mathrm{Definite}\left(F\,k=0..\mathrm{\infty }\right)\mathbf{assuming}\mathrm{abs}\left(z\right)\le 1$

$\sum _{k=0}^{\mathrm{\infty }}\frac{2^{2k}\mathrm{\Gamma }\left(-n+k\right)\mathrm{\Gamma }\left(k+n\right)z^k}{\sqrt{\mathrm{\pi }}\mathrm{\Gamma }\left(2k+1\right)}=-\frac{\sqrt{\mathrm{\pi }}\cos \left(2n\arcsin \left(\sqrt{z}\right)\right)\csc \left(\mathrm{\pi }n\right)}{n}$

$F≔\frac{\mathrm{binomial}\left(2k-3\,k\right)}{4^k}$

$F≔\frac{\left(\genfrac{}{}{0ex}{}{2k-3}{k}\right)}{4^k}$

$\mathrm{Sum}\left(F\,k=0..n\right)=\mathrm{Definite}\left(F\,k=0..n\,\mathrm{parametric}\right)$

$\sum _{k=0}^n\frac{\left(\genfrac{}{}{0ex}{}{2k-3}{k}\right)}{4^k}=\left\{\begin{array}{cc}\frac{\left(\genfrac{}{}{0ex}{}{2n-1}{n}\right)\left(4+2n\right)}{4^{n+1}}& n\le 0\\ \frac{3}{4}& n=1\\ \frac{\left(\genfrac{}{}{0ex}{}{2n-1}{n}\right)\left(4+2n\right)}{4^{n+1}}+\frac{3}{8}& 2\le n\end{array}\right.$

$\mathrm{Definite}\left(\frac{1}{k}\,k=a..b\right)$

$\sum _{k=a}^b\frac{1}{k}$

$\mathrm{Definite}\left(\frac{1}{k}\,k=a..b\,\mathrm{parametric}\right)$

$\left\{\begin{array}{cc}0& a=b+1\\ \mathrm{\Psi }\left(b+1\right)-\mathrm{\Psi }\left(a\right)& 1\le a\mathbf{and}0\le b\\ \mathrm{\Psi }\left(-b\right)-\mathrm{\Psi }\left(1-a\right)& a\le 0\mathbf{and}b\le \mathrm{-1}\\ \mathrm{FAIL}& \mathrm{otherwise}\end{array}\right.$

$F≔\frac{t^2+1}{t^3-5t+2}$

$F≔\frac{t^2+1}{t^3-5t+2}$

$\mathrm{Definite}\left(F\,t=\mathrm{RootOf}\left(x^5+x+1\right)\right)$

$-\frac{269}{833}$

Egorychev, G.P. "Integral Representation and the Computation of Combinatorial Sums." Novosibirsk, Nauka. (1977). (in Russian); English: Translations of Mathematical Monographs. Vol. 59. American Mathematical Society. (1984).

Roach, K. "Hypergeometric Function Representations." Proceedings ISSAC 1996, pp. 301-308. New York: ACM Press, 1996.

van Hoeij, M. "Finite Singularities and Hypergeometric Solutions of Linear Recurrence Equations." Journal of Pure and Applied Algebra. Vol. 139. (1999): 109-131.

Zeilberger, D. "The Method of Creative Telescoping." Journal of Symbolic Computing. Vol. 11. (1991): 195-204.