When I simplify the following two (to my mind identical) expressions, I get different outputs. But why?

s1:=exp(ln(((mu)^(2))/(sqrt(((sigma^(2)))+(mu)^(2))))+(1-phi)*ln((sqrt((((sigma^(2))))+((mu)^(2))))/(mu)));

simplify(s1);

and now just replacing sigma^2 by sigma2 gives:

s2:=exp(ln(((mu)^(2))/(sqrt(((sigma2)))+(mu)^(2))))+(1-phi)*ln((sqrt((((sigma2))))+((mu)^(2))))/(mu)));

simplify(s2);

I tried to solve the following nonlinear equation system:

with(Statistics):

Y9 := RandomVariable(LogNormal(a, b)):

solve({mu=Mean(Y9),(sigma)^(2)=Variance(Y9)},{a,b}, UseAssumptions) assuming a::real, mu::real, sigma>0, b>0;

The problem is that the conditions, which discriminate between the cases of the piecewise solution are puzzling, not to say impossible to meet.

E.g. at the end of the first row it says:

I tried to skip some iterations of a do loop, by using the following code:

for i from 1 to 19 do
if i=8 or i=16 then
  next
end if:

...

Unfortunately Maple just accounts for the i=8 but not for the second condition i=16.

How may I specify more than one condiions in an if statement?

I´d like to compute some sums where x ranges to infinity.

How may I compute that in Maple?

sum(ln(x)/((x)^(2)+3*x+2.),x=1..infinity):

didn´t work.

add(ln(x)/((x)^(2)+3*x+2.),x=1..10000):

is possible and though I know the limit of the expression

limit(ln(x)/((x)^(2)+3*x+2),x=infinity) is known and zero, I have no idea of how to compute the sum to infinity from the first place.

When I execute a do loop and an invalid output shows up, Maple breaks the loop.

Is there a way to tell Maple to show me, where the error occured, but nonetheless execute the loop to the very end?