kill(all)$

maxima:sum infinity ( = value ?),version 1.0,part 2
peter.vlasschaert@gmail.com,08/12/2020

see:maxima:sum infinity ( = value ?),version 1.0,part 1

packages  , option

load(simplify_sum)$
load(lrats)$ratprint:false$

Given : infinite sum

p1:S='sum(3^n/((n+3)*(n+4)*n!),n,0,∞);

We need function : no faculty , denominator,
sin          → faculty
cos         →faculty
log(x+1)   →no faculty
exp(x)      →faculty
-------------------------------------------------------------

use : menu:Calculus:Get Series

1e) step

q1:niceindices(powerseries(sin(x), x, 0))$
q1a:sin(x) =q1;
q2:niceindices(powerseries(cos(x), x, 0))$
q2a:sin(x) =q2;
q3:niceindices(powerseries(log(1-x), x, 0))$
q3a:(log(1-x) =q3); /*math: ln(1-x)=log(1-x) */
q4:niceindices(powerseries(exp(x), x, 0))$
q4a:(exp(x) =q4); 

2e) step ,need falculty in denominator:we use  exp(x)

→ denominator , smallest factor (n+3)

assume(n>0);

qint(expr,xx):=integrate(expr,xx);

p1a:integrate(x^(n),x);

p1a1:qint(qint(qint(x^n,x),x),x);

what about:exp(x)*x^m

assume(m>0)$

p1a1:intosum(q4a*x^m);

3e) step: m, so that we get n+3 ,after integration,denominator

p2a:subst([m=2],p1a1);

p2a1:'qint(lhs(p2a),x)='qint(rhs(p2a),x)+c1;

p2a2:ev(p2a1,nouns);

4e) step : find c1, put x=0, first integration

p3a:reverse(subst([x=0],p2a2));

p3a1:subst([p3a],p2a2);

p3a2:p3a1-rhs(p3a);

5e) step : find c2, put x=0, first integration

p4a:qint(rhs(p3a2),x)=qint(lhs(p3a2),x)+c2;

p4b:solve(subst([x=0],p4a),c2);

p4c:subst(p4b,p4a);

6e) step : put x=3,p4c

p5:subst([x=3],p4c);

p5a:subst([i=n],p5);

? check: with maxima
 -->    /*lhs(p3a1)=simplify_sum(lhs(p3a1))*/;7e) step :representation : find factor , needed to find S,see :above.

p6:args(rhs(p1))=args(lhs(p5a));

p6a:lhs(p6)[1]/rhs(p6)[1];

8e) final result: S,use p6a

p7:intosum(p6a*p5a);

p7a:lhs(p7)=float(rhs(p7));

? check: with maxima

p8:lhs(p7)=simplify_sum(lhs(p7));

p8a:float(p8);