@H-R 

A shorter formatted proof:

by definition, for any branch.

@Markiyan Hirnyk 

I answer with great pleasure when the person who asked the question does not understand the solution. And I did it each time. Now I do not know what to think.

Is it not clear that

-RootOf(_Z^2*s-s+1)*s/(s-1)

equals

+-  sqrt(s*(s-1))/(s-1)

?

Must I elaborate? Must I prove that Maple is correct? Are you serious?

@Markiyan Hirnyk 

Why mine and not yours?

But more important, why must we waste our time for such a simple problem?

Just apply allvalues as you did in your post.
Or even simpler. use the definition of RootOf.

@Markiyan Hirnyk

@Markiyan Hirnyk 

It was a simple typo. The principle was obviously correct.

[a = s/r, b = -r/s, c = r, r^2-s^2+s=0 ];
eliminate(%,r);

solve(%[2], {a,b,c});

which is the same as the original.

@Markiyan Hirnyk 

You are as usual ready to object without reason.

The correct call is

solve({a*c-s, b*s+c, -c*s^2+c^2+s}, {a,b,c});

which gives the expected result.

@Nikol 

F:= unapply( rsolve({f(n) = .5*f(n-1)+.5*f(n+1)}, f(n)),  n);
solve( {F(a)=0, F(0)=1}, {f(0),f(1)} );
simplify(eval(F(n),%));

@Preben Alsholm 

Ok, ler's call it semi-bug or quasi-bug :-)

But unfortunately

rsolve(f(n+1)=f(n)+c, f(n));

Congratuations, you have found a bug!

The correct answer should be of course f(n) = 1 - n/6.

It seems that rsolve fails for arithmetic progressions!

E.g. also:

rsolve(f(n+1)=f(n)+c, f(n));

Strange! The bug seems to be new.

@Axel Vogt 

0.002744071672397914633171931780975747758...

And symbolically:

@Markiyan Hirnyk 

OK, can you obtain more than 5 corect digits this way?

Edit

exact = 4.0148672762003073023634...*10^9

@Markiyan Hirnyk 

It works indeed but with very poor accuracy: only 3 correct digits in this case.

For this, f must be a formal powerseries (not a series in the Maple sense, which is a truncation, containing O(...)).

So,

f:=convert(exp(k*t)*cos(w*t),FPS,t,n);

g:=convert(f/exp(k*t),FPS,t);

After several conversions, it it possible to simplify g to
But probably for more complicated functions this will not work.

@Preben Alsholm 

This behavior of the roots is known.

If the coefficients are independently and identically distributed with a mean of zero,
the complex roots are on or close to the unit circle.

See:
https://en.wikipedia.org/wiki/Properties_of_polynomial_roots