Download lim_ser-vv.mw

Maple can guess the result:

s:=sum( arctan(2/n^2), n=1 .. infinity);
ans:=identify(evalf(s));
evalf[200](s - ans);  # check 200 digits, --> 0

PS. Why don't you express the series in Maple, i.e.  sum( arctan(2/n^2), n=1 .. infinity) ?

PS2. Of course this is not a proof.  It is easy to obtain one in Maple, but only if the user knows it mathematically.
I omit it because you probably know it.

Maple cannot compute it, but it's easy with Stoltz-Cesaro theorem:

a:=n -> 2^n/n:
limit(a(n) / (a(n)-a(n-1)), n=infinity);

                    2

Download INT1-vv.mw

Try  (*.mw, not *.mpl):

interface(worksheetdir);

 1/(1 - x*y) = Sum( (x*y)^k, k=0 .. infinity );
Applying the double integral over (0,1)^2 implies (by monotone convergence):
 Int( 1/(1 - x*y), x=0..1, y=0..1 ) = Sum( 1/k^2, k =1 .. infinity );
==>  Int( 1/(1 - x*y), x=0..1, y=0..1 ) = Pi^2 / 6.

You have to use LinearAlgebra[Modular]:

restart
A:=Matrix(9,[
0,0,0,-1,1,-1,-1,0,0,
0,-1,0,1,-1,1,0,-1,-1,
0,0,-1,0,0,0,1,1,0,
0,1,0,-1,-1,1,-1,-1,0,
0,0,0,0,0,-1,1,0,-1,
0,0,0,0,0,-1,0,1,1,
0,0,1,-1,1,1,-1,-1,0,
0,0,0,0,0,1,1,-1,0,
0,0,0,0,0,0,0,0,0], datatype=integer[8]):
with(LinearAlgebra[Modular]):
Rank(3,A);
#                               7
B := Copy(3,A):
RowReduce(3,B,9,9,9,'det',0,'rank',0,0,true): B;

Edit. To obtain a basis for the nullspace explicitly, it is better to use:

with(LinearAlgebra[Modular]):
A1 := Copy(3,A):
r:=MatBasis(3, A1, 9, true): r,A1;
colonbasis, nullspacebasis:= A1[1..r, ..], A1[r+1.., ..];
colonbasis . nullspacebasis^+  mod 3; # check (==> 0 matrix)

OK, I'll take Z(t) = exp(int(y(u),u=0..t))

restart
eq1 := diff(x(t), t)-(1/6)*(6*x(t)^3*y(t)+(2*y(t)^2-2)*x(t)^2+3*y(t)*(z(t)-2)*x(t)-2*y(t)^2+2)*sqrt(3) = 0;
eq2 := diff(y(t), t)-(1/6)*(y(t)-1)*sqrt(3)*(y(t)+1)*(6*x(t)^2+2*y(t)*x(t)+3*z(t)-2) = 0;
eq3 := diff(z(t), t)-(1/3)*z(t)*sqrt(3)*(6*y(t)*x(t)^2+2*x(t)*y(t)^2+3*z(t)*y(t)-2*x(t)-3*y(t)) = 0;
eq4 := diff(s(t), t)-(1/3)*(y(t)-1)*sqrt(3)*(y(t)+1)*(6*x(t)^2+2*y(t)*x(t)+3*z(t)-2)*y(t) = 0;
eq5 := diff(Q(t), t)+sqrt(3)*((4/3-z(t)-2*x(t)^2)*y(t)+(2/3*(1-y(t)^2))*x(t))*Q(t)+2*((2/3)*y(t)+x(t))*P(t)/sqrt(3)-(2/3)*R(t) = 0;   
eq6 := diff(P(t), t)+(sqrt(3)*(1-z(t)-2*x(t)^2)*y(t)+2*(2-y(t)^2)*x(t)/sqrt(3))*P(t)+2*sqrt(3)*x(t)*Q(t)+(1/2)*R(t) = 0;       
eq7 := diff(R(t), t)-sqrt(3)*((3*x(t)^2+(3/2)*z(t)+4)*y(t)-(1/3*(1-3*y(t)^2))*x(t))*R(t)-z(t)*P(t) = 0;
eq8 := diff(T(t), t)-sqrt(3)*((6*x(t)^2+3*z(t)+1)*y(t)-(2*(1-y(t)^2))*x(t))*T(t)-(1-y(t)^2)*(Q(t)-(2/3)*P(t)) = 0;   

eq9 := diff(Z(t), t) = Z(t)*y(t);
syst := eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8, eq9;
ics := x(0) = .1, y(0) = .9, z(0) = .1, s(0) = .2, Q(0) = .1, P(0) = .9, R(0) = .1, T(0) = .2,  Z(0)=1;
sol1 := dsolve({ics, syst}, {P(t), Q(t), R(t), T(t), s(t), x(t), y(t), z(t), Z(t)}, type = numeric, output = operator);
sol1(0.5);
plots:-odeplot(sol1, [[t, y(t)], [t,Z(t)]], t=0..1, title="y and Z");
plots:-odeplot(sol1, [Z(t), y(t)], t=0..1, title="y vs Z");

Take the origin at A. Then the area a(t) of the triangle ABC is the absolute value of the determinant of <B,C>, so a(t) = |m t + n|.

We have a(0) = 2, a(5) = 3, a(10) = ?.
a(0)=2 implies n=±2, and a(5)=3 implies 5m+n=±3 so, a(10) = |10m+n| = |-n±6| = |n±6| ∈ {4,8}.

restart;
A := t->[a,sA+vA*t]: B := t->[b,sB+vB*t]: C := t->[c,sC+vC*t]:
area:= t->abs(LinearAlgebra:-Determinant(Matrix([[A(t)[],1],[B(t)[],1],[C(t)[],1]])))/2:
sols := solve([area(0)=2, area(5)=3, area(10)=AREA]):
AREA in {seq}( eval(AREA, sol), sol=sols);

                AREA  ∈  {4, 8}

(so, 2 solutions, AREA=4 and AREA=8).
Nice problem BTW.

p:=unapply(LinearAlgebra:-CharacteristicPolynomial(A,x), x):
[seq]( simplify(p(evstheory[i])), i=1 .. L^2); nops(%);

        [0, 0, ... , 0]
        81

but very SLOW!

Just the limit (it can be obtained at once).
Let A,B,C be consecutive vertices and M,N the midpoints of AB, BC.
For n --> oo, angle(ABC) --> Pi, so AC --> 2AB = 2 and MN = AC /2 --> 1.

Nice problem.

Conside the cube having lattice vertices  (x,y,z), with x,y,z in {0,1}.
We decompose the cube in six pyramids having a vertex in P(a,b,c) 
and the bases formed by the faces of the cube.
Denote by d1,...,d6 the distances from P to the six faces.
Then d1 + ... + d6 = 1+1+1 = 3 and
d1/1 = d2/3 = ... = d6/6 = (d1+ ... +d6)/(1+ ... +6) = 3/21 = 1/7.
So, {d1,...,d6} = {1/7, 2/7, 3/7, 4/7, 5/7, 6/7}.
Then the desired vertex is P(1/7,2/7,3/7}  
(because (1-1/7, 1-2/7, 1-3/7}= {4/7, 5/7, 6/7}).

P.S. All your proposed problems are nice but not very suitable to be solved
with Maple. Nice problems for a CAS can be found in:

Eilers S., Johansen R. - Introduction to Experimental Mathematics. CUP, 2017.

x*t is not a subexpression (operand of any level) in L. Use:

algsubs(x*t = Y, L)

dsolve( diff(y(x),x) = (x+y(x))/(y(x) - 3));

      -(1/2)*ln(-((x+3)^2-(x+3)*(3-y(x))-(3-y(x))^2)/(x+3)^2)-(1/5)*sqrt(5)*arctanh((9-2*y(x)+x)*sqrt(5)/(5*(x+3)))-ln(x+3)-_C1 = 0

You may append explicit, for an explicit solution.