I did not mean cancellation through sum a series, but 2 (or 3) terms ... but I am not sure what Maple does for the task.

However I bet that it does not try to evaluate the series if |z| ~ 1

The shown result indicates a numerical error which typically occurs through cancellation by adding floats.

The reason here might be: for z ~ 1 the system applies a transformation to reduce to series at zero, for 2F1 it might be 1 - 1/z, giving 2 terms. And if those are of same magnitude their trailing decimal places will cancel. MMA may use different transforms.

Yes, I already know that Maple does not solve it - else I would have stated that. But I thought you want to re-find a proof.

At least at http://functions.wolfram.com/EllipticIntegrals/EllipticPi/03/ShowAll.html one finds an "answer" for the last reduction. I am not used to MMA notation (and the explanations there seem to assume that, the documentation is not good) and it might be that there are typos / errors.

Anyway: this is a route to dig.

It is the reduction to 1 elliptic function ...

The function advisor says that it does not know an asymptotic expansion (besides that here there is some 'diagonal' situation)

I think the most simple form for your task would be to prove: Limit(EllipticPi(r,r)*(1-r), r = 1, left) = Pi/4 or likewise in a formulation EllipticPi(1-epsilon,1-epsilon)*epsilon for epsilon towards 0.

If I remember an older discussion about that: the credentials are stored in cookies with expiry = 1 year. Without additional work a user will continue to have access if s/he does not delete cookies.

And I doubt that captchas will do: work is very cheap in many countries.

BTW: IIRC "deleting" only prevents to show those posts, and I assume that an admin will check it (in case of errors for example)

@Markiyan Hirnyk 

Usually I use solve and and allvalues in case of need. For Maple 18 I get the numerical values, only (picking Reals is possible, I always avoid command options for that)

Having no MMA I can ask the online version, but it only returns the real but exact solution. So not sure what MMA answers for older versions.

That does not answer your question, I know.

@ecterrab 

I understand my problem now (taking m=0 for simplicity).

Sum(A(i)*B(j), i+j=0) means summing over all integers s.th. i+j=0 (not
regarding an order) while in Maple only non-negative integers are meant.

@ecterrab 

I am a bit uncertain as well and find the notation not clear.

For me j+k = constant ( = some m, fixed, m = ...) is a 'line' in Z^2. 
And in Math I would not use it (except it is meant as such).

  a^j*b^k;   isolate(j+k=m, j);   eval(%%, %);   Sum(%, k=-infinity.. infinity);

I do not understand what is meant.

@necron are you sure that the exp at erfc is correct - linear in y?

Rouben, sorry for being sloppy: I did not mean to compute the winding number (your approach is fine for me) by integration. But it is sometimes easy to fix some points and *use that* curve to integrate. Carl's procedure provides it without doing the curve by hand (if being feed reasonable).

No actual solution: g1 lives only on an interval of length Pi

And why it should give "constant figure"? I get 936.726407954295, 72.9285903589035, 43.2708943259480, 18.8478319773527 for t = Pi, 10, 5*Pi, 20 for epsilon = 1e-6, Digits=15 using some equivalent way (I would say).

But very slow, yes: a very (!) lengthy result for g5 using expensive Fresnel functions (lenght = 222301). Already that makes it error prone for numerical calculations

@Carl Love Thx. Sometimes it is or might be convenient to have such in the plane for a Cauchy integral (and split the integral piecewise)

Edited to add: I understand that the user is responsible to provide an according sequence of points ... it 'only' connects the points given in the list according to its ordering - yes?

Question aside: Is it clear from construction that the curve is closed with winding number 1?

May be you mean something like this

B:=int(t^(a-1)*(1-t)^(b-1),t=0..x);
eval(B,{x=sqrt(t),a=1/5,b=1/2});
convert(%, cos): normal(%): combine(%):
beta:= unapply(%, t);

plot(beta(t), t=0..1);
plottools[reflect](%,[[0,0],[1,1]]);

It would be better if you first write it down in mathematical notation to see what you actually want. And then try to formulate it in Maple's syntax.