kill(all)$

maxima:sum infinity ( = value ?),version 1.0,part 1
peter.vlasschaert@gmail.com,08/12/2020

packages  , option

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

Given : infinite sum

p1:S='sum(1/(n*(n+1)*3^n),n,1,∞);

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, we want , n+1,into dominator. 'integration'

p1a:(assume(n>0),integrate(x^n,x));

p1a1:integrate(p1a,x);

we get product of factors in the dominator : (n+1)*(n+2),after integration 
again

forget(n>0)$/*forget predicate for n */

→ :integrate(x^n,x),don't use "forget" for the  'assume' statement 
answer : Is n equal to -1 ?, question : Maxima ask to you.
→ : option , change boolean , 'see : ratprint:false' .
answer : ratprint;true   

→integration both sides  ( don't forget integration constant=c1)

p2a:'integrate(lhs(q3a),x)='integrate(rhs(q3a),x)+c1;

p2a1:ev(lhs(p2a),nouns) = ev(rhs(p2a),nouns)+c1;

find c1:

p2a1a:subst([x=0],p2a1);

p2a1b:solve(p2a1a,c1)[1];

3e) step x=1/3

p2a2:intosum(subst([x=1/3,i=n,p2a1b],p2a1));

p2a2a:(-1)*(p2a2-1);

sum((-1)^n,n,1,inf),simpsum; always no convergence,interchange sign

representation: 'p2a2'

p2a3:intosum(lhs(p2a2a)=rhs(p2a2a));

use :rhs(S), show

p2a4:args(rhs(p1))=args(rhs(p2a3));

→ rhs

p2a5:rhs(p2a4)[1]/lhs(p2a4)[1];

to lhs

p2a6:1/p2a5;

4e) step: 'use p2a6*p2a3

p3a:intosum(p2a6*(p2a2a));

p3a1:reverse(distrib(p3a));

? check: with maxima

p4a:lhs(p3a1)=simplify_sum(lhs(p3a1));