Function: sumalt
Section: sums
C-Name: sumalt0
Prototype: V=GED0,L,p
Help: sumalt(X=a,expr,{flag=0}): Cohen-Villegas-Zagier's acceleration of
 alternating series expr, X starting at a. flag is optional, and can be 0:
 default, or 1: uses a slightly different method using Zagier's polynomials.
Wrapper: (,G)
Description:
  (gen,gen,?0):gen:prec sumalt(${2 cookie}, ${2 wrapper}, $1, prec)
  (gen,gen,1):gen:prec sumalt2(${2 cookie}, ${2 wrapper}, $1, prec)
Doc: numerical summation of the series \var{expr}, which should be an
 \idx{alternating series}, the formal variable $X$ starting at $a$. Use an
 algorithm of Cohen, Villegas and Zagier (\emph{Experiment. Math.} {\bf 9}
 (2000), no.~1, 3--12).

 If $\fl=1$, use a variant with slightly different polynomials. Sometimes
 faster.

 The routine is heuristic and a rigorous proof assumes that the values of
 \var{expr} are the moments of a positive measure on $[0,1]$. Divergent
 alternating series can sometimes be summed by this method, as well as series
 which are not exactly alternating (see for example
 \secref{se:user_defined}). It should be used to try and guess the value of
 an infinite sum. (However, see the example at the end of
 \secref{se:userfundef}.)

 If the series already converges geometrically,
 \tet{suminf} is often a better choice:
 \bprog
 ? \p28
 ? sumalt(i = 1, -(-1)^i / i)  - log(2)
 time = 0 ms.
 %1 = -2.524354897 E-29
 ? suminf(i = 1, -(-1)^i / i)   \\@com Had to hit <C-C>
   ***   at top-level: suminf(i=1,-(-1)^i/i)
   ***                                ^------
   *** suminf: user interrupt after 10min, 20,100 ms.
 ? \p1000
 ? sumalt(i = 1, -(-1)^i / i)  - log(2)
 time = 90 ms.
 %2 = 4.459597722 E-1002

 ? sumalt(i = 0, (-1)^i / i!) - exp(-1)
 time = 670 ms.
 %3 = -4.03698781490633483156497361352190615794353338591897830587 E-944
 ? suminf(i = 0, (-1)^i / i!) - exp(-1)
 time = 110 ms.
 %4 = -8.39147638 E-1000   \\ @com faster and more accurate
 @eprog

 \synt{sumalt}{void *E, GEN (*eval)(void*,GEN),GEN a,long prec}. Also
 available is \tet{sumalt2} with the same arguments ($\fl = 1$).