How can I determine inf and sup of the set:
$$\ {{p*\ln(p) + q*\ln(q):~~~~~~ p,q>0, ~~~p+q=1}} $$
I have modified to:
$$\ (1-q)\ln(1-q)+q\ln(q) = \ln(1-q)^{(1-q)} + \ln(q)^q = \ln((1-q)^{(1-q)}*(q)^q) = \ln(\frac{1-q}{(1-q)^q}*q^q) = \\ \ln((1-q)*(\frac{q}{1-q})^q)$$
I tried to determine the radius of convergence, using the last equation as the formula for power series, however, in $\ \lim|( \ln((1-q)*(\frac{q}{1-q})^q)/ \ln((q)*(\frac{q+1}{q})^{q+1}))| $ I got nothing relevant. Do I have to use dfferentiation or Taylor series? I would appreciate any explanation.
Keep the form
$$f(q)=q \ln(q) + (1-q) \ln(1-q)$$
(this function is a version of the so-called entropy function, more precisely its opposite).
You will find that its derivative is
$$f'(q)=\ln\left(\frac{q}{1-q}\right)$$
As $$\frac{q}{1-q}\begin{cases}<1& \ \ if \ \ 0<q<\frac12\\>1& \ \ if \ \ \frac12<q<1\end{cases}$$
we have $$f'(q)\begin{cases}<0& \ \ if \ \ 0<q<\frac12\\>0& \ \ if \ \ \frac12<q<1\end{cases}.$$
Therefore : $$f \begin{cases}\downarrow & \ \ if \ \ 0<q<\frac12\\ \uparrow & \ \ if \ \ \frac12<q<1\end{cases}$$
as attested by the blue curve below.
Fig. 1 : The curve of $f$ (in blue) and its approximation by the parabolic curve (in red) of function $g(q)=-\log(2)+2(q-0.5)^2$ (coming from Taylor expansion).
Therefore $f$ is minimal for $q=1/2$ with $f(1/2)=-\ln(2) \approx -0.693...$
Its sup value $0$ is obtained at the boundary, either for $q=0$ or $q=1$.
In fact, you must know a classical limit $\lim_{q \to 0}q \ln q =0$ in order to be able to say that :
$$\lim_{q \to 0}f(q)=\lim_{q \to 0} q\ln(q)+\lim_{q \to 0} (1-q)\ln(1-q)=0+1 \ln(1)=0.$$
As a matter of conclusion, a funny fact concerning this curve (or more exactly its version with $\ln$ replaced by $\log_{10}$) can be found there : http://www.mi.sanu.ac.rs/vismath/miyadera/999/node4.html cited by one of the big contributors to this site, Gerry Myerson.