@FlamingFox 

In the first procedure (=J) the condition 1<n<k  is not correct in 1D math and not needed in 2D; it should be simply n<k  as in J.
In the last return you have too many ")".

@FlamingFox 

Then it would be nice to post your corrected procedure.

@Kitonum 

Of course, k>1. I have not included a check for this.
Or, one may insert
if k=1 then return n-1 fi;
but I prefer to keep it as it is.
 

@Markiyan Hirnyk 

It is possible to compute by hand stating from

sum(k*z^k/(k^2+p^2+k), k = 1 .. infinity)

for |z|<1, and expanding k/(k^2+p^2+k) in partial fractions. Then LerchPhi appears.
But it's very time consumming.

For example,

sum( (a^2*k+sin(a))^3/(1+a^2*k+a^4*k^3), k=1..infinity) assuming real;
returns the generic result infinity, which is valid for a<>0.
With option parametric I would expect the result:
piecewise(a=0, 0, infinity);
Unfortunately it is infinity.

@Markiyan Hirnyk 

@Kitonum 

A generic answer should be valid for most cases (usually except a finite set).
'parametric' should take care of those exceptions.

My impression is that 'parametric' works only for simple situations.

@Earl

There are examples in the Programming Guide.
To construct complex types, see ?type,structure
You can also inspect the first lines of code of Maple commands and compare with their description in the help page.
E.g.
showstat(plottools:-disk); # a simple one
showstat(Groebner:-Basis); # a complex one

But for a standard user procedure the old fashion way is usually enough (even if not "elegant"): simple types for parameters and then check the parameters in the body of the procedure.

@miguelbravo 

The difference LHS-RHS  is very small  (about 10^(-43)) for Digits=40.

It becomes 10^(-80)  for Digits=80  and 10^(-120)  for Digits=120.

So,  RHS-LHS is practically 0. It seems that your constants are not accurate enough or the model is not reliable.
 

@jamunoz 

I think I have explained that it's not possible.
For a given t you cant find a unique pair (t1,t2). E.g. what are t1 and t2  if t = 7?
Your problem is a mathematical nonsense. If it comes from another problem you want to solve then please show that problem and we could try to help.

@Markiyan Hirnyk 

But it's not a bug, it's just a notation (it appears in all the help  examples). Maybe it could be explained better (with that f@phi).
A mathematician would use F(u,v)  for (f@phi)(u,v) but Maple prefers not to use another name; and actually many engineers do the same.

@Markiyan Hirnyk

No, Maple is right. Keep in mind that f(u,v) is actually (f@phi)(u,v)  where phi is the diffeomorphism but Maple uses the same notation.

@Markiyan Hirnyk 

Yes, I also think that `a` should not appear. But if we add params = [a] then it is OK. It is true that in this case the new variables can be omitted. Maybe you are right that "it's not simple" :-)

P.S. I don't understand your PS

@jamunoz 

f must have either one or two variables.

Now, say you have a function f(t) of one variable.
You can define a function g of two variables:

g:=(t1,t2) -> f(t1/(k*m) + t2/(k*m^2));
# Let us compute:
d:=diff(g(t1,t2),t1)*k*m + diff(g(t1,t2),t2)*k*m^2;

#  Observe the "2";  this is the correct result.

# If you want to appear "t" you can use:

eq := t1/(k*m)+t2/(k*m^2) = t;
subs(eq,d);

You don't need and cannot use dchange for this.
 

@jamunoz

It is not a Maple limitation. It's maths!
You will have to explain what you are trying to do.

@Markiyan Hirnyk 

If the change of variables is e.g.  {x=u+v, y=u-v}, then the names of the new vars {u,v} are not necessary.

But for {x=u+v, y=u-v+a}, you must declare {u,v}. It's simple.