If  n = 1  then you get  0  in the denominator of  c(n) , so I chose  n = 2 .. infinity . The method you use to find the convergence region is guaranteed to only give an open interval. Therefore, we must separately investigate the convergence at the ends of this interval:

 

restart;
c:=n-> 1/(n^2-1);
u:=n->c(n)*(x-2)^n; # The function of the general term of the series
S:=Sum(u(n), n=2..infinity);
limit(abs(u(n+1)/u(n)), n=infinity); # We use the ratio test (d'Alembert test) for a series of absolute values
solve(%<1); # We get the (open) convergence interval
x1,x2:=op(op(1,%)),op(op(2,%)); # The left and right ends of the convergence interval 

value(eval(S, x=x1)); # The series converges at the left end
value(eval(S, x=x2)); # The series converges at the right end

Edit.

We will denote  U[n]=u(x[n])  for all  n=1..N

Download Sys.mw

Of course, the indefinite integral is calculated up to an additive constant. Here's a way to get the desired result.

Download QuestionInt_new.mw

You can use  combinat:-permute  for this:

restart;
S:=[1,2,3]:
combinat:-permute(map(`$`,S, 2), 2);

                   [[1, 1], [1, 2], [1, 3], [2, 1], [2, 2], [2, 3], [3, 1], [3, 2], [3, 3]]

or a little shorter

combinat:-permute(`$`~(S, 2), 2);

Edit. If you use  the  Iterator  package then the  CartesianProduct  command can be used.

Download simplify.mw

Download intersect_curve_new.mw

restart;
f:=unapply(int(diff(Y1(x), x), x)+C1, x);
solve({Y1(0) = f(0)}, {C1});

                                    f := x -> Y1(x) + C1
                                           {C1 = 0}

For drawing arrows see help on  plots:-arrow

restart;
with(plots): with(plottools):
S:=seq(seq(disk([x,y], 0.05, color=blue), x=-1..6), y=-1..3):
T:=textplot([[0,0,"(0,0)",align=[left,below]],[1,1,"(1,1)",align=[right,above]],[2,1,"(2,1)",align=[right,above]]], font=[times,18]):
display(S,T, scaling=constrained, size=[800,500],axes=none);

Two-argument  arctan(y,x)  returns the polar angle  phi  of a point  A(x,y) in the range  -Pi < phi <= Pi

Examples:

restart;
arctan(1,sqrt(3)), arctan(-1,sqrt(3)), arctan(-1,-sqrt(3)), arctan(0,-1);  

See help on  ?arctan  for details  .

Example:

combs := combinat:-choose({a, b, c, d, e}, 3);

     combs := {{a, b, c}, {a, b, d}, {a, b, e}, {a, c, d}, {a, c, e}, {a, d, e}, {b, c, d}, {b, c, e}, {b, d, e}, {c, d, e}}

What functions are you talking about? Inside contourplot command, you simply calculate the sum of the values of one function  BesselJ  in a finite number of points, and skip one  sum command, since you have a double sum:

restart;
sum(sum(BesselJ(n,r),n=0..5),r=-10..10);

                 1+2*BesselJ(0, 10)+2*BesselJ(2, 10)+2*BesselJ(4, 10)+2*BesselJ(0, 9)+2*BesselJ(2, 9)+2*BesselJ(4, 9)+2*BesselJ(0, 8)+2*BesselJ(2, 8)+2*BesselJ(4, 8)+2*BesselJ(0, 7)+2*BesselJ(2, 7)+2*BesselJ(4, 7)+2*BesselJ(0, 6)+2*BesselJ(2, 6)+2*BesselJ(4, 6)+2*BesselJ(0, 5)+2*BesselJ(2, 5)+2*BesselJ(4, 5)+2*BesselJ(0, 4)+2*BesselJ(2, 4)+2*BesselJ(4, 4)+2*BesselJ(0, 3)+2*BesselJ(2, 3)+2*BesselJ(4, 3)+2*BesselJ(0, 2)+2*BesselJ(2, 2)+2*BesselJ(4, 2)+2*BesselJ(0, 1)+2*BesselJ(2, 1)+2*BesselJ(4, 1)

Of course, you can plot  a  contourplot for the function of two variables  BesselJ(n,r)  in appropriate ranges for n and r without any problem:

plots:-contourplot(BesselJ(n,r), n=0..5, r=-10..10, grid=[200,200]);

Maple finds both real and complex solutions. The  select  command helps to select real roots:
 

Edit. Of course, you can immediately get the solution (as float number) you want using  fsolve  command:

sol := fsolve({F, G}, {x, y});

                                       sol := {x = 0., y = -2.000000000}

Download solve.mw

@Zeineb  What you want to calculate is called a (matrix) norm. Carl showed one way (the  infinity-norm). But there are other formulas for calculating a norm, and Maple has a suitable command. See help on  ?LinearAlgebra:-Norm .

Obviously, the function  g  depends not only on x, but also on p , so  g(x,p)=max{h(x,p),f(x,p) . It is useful to use graphic illustration to obtain this dependence. First, we plot a red line  x-p=x^2+p*x . It is easy to check that above this line (pink area) we have  g(x,p)=x^2+p*x, and below (blue area)  g(x,p)=x-p

Download int.mw

Here's another solution that is slightly more cumbersome but gives more compact results:

Download int1.mw

Edit.

Download RM.mw