Does the series $S=\sum\limits_{n=1}^\infty a_n\log(a_n)$ with $\sum\limits_{n=1}^\infty a_n=1$ and $a_n\in(0,1]$ always converge? Given the following sum:
\begin{align}
S=\sum\limits_{n=1}^\infty a_n\log(a_n)
\end{align}
And this "contract":
\begin{align}
\sum\limits_{n=1}^\infty a_n&=1 \\
a_n&\in(0,1]
\end{align}
Does the sum $S$ always converge? I tried a few coefficients (e.g. $a_n=\frac{6}{\pi^2 n^2}$ or $a_n=2^{-n}$), but i only found converging values.
Thank you very much
 A: Short answer: no.
Long answer: we may consider that by Cauchy's condensation test the series $\sum_{n\geq 1}\frac{1}{n\log^2(n+1)}$ is convergent to a positive constant $L$. Let $a_n = \frac{1}{L\,n\log^2(n+1)}$. Then $\{a_n\}_{n\geq 1}$ obviously fulfills the constraints $a_n\in(0,1]$ and $\sum a_n=1$, but
$$ \left|a_n \log(a_n)\right| \geq \frac{C}{n\log(n+1)} $$
hence, always by Cauchy's condensation test, the series $\sum_{n\geq 1}a_n\log(a_n)$ is not converging.
A: We build a sequence $(a_n)$ that is a counterexample:
For each $n\geq 1$ pick a $w$ such that $w \log(\frac{1}{2^nw})\frac{1}{2^n}\geq 1$, let $w=b(n)$.
Let the first $b(1)$ terms of $(a_n)$ be $\frac{1}{b(1)2^1}$
Let the next $b(2)$ terms of $(a_n)$ be $\frac{1}{b(2)2^2}$
Let the next $b(3)$ terms of $(a_n)$ be $\frac{1}{b(3)2^3}$ and so on.
Clearly the sum of $a_n$ is $\frac{1}{2}+\frac{1}{2^2}\dots =1$.
However the sum of each block of $a_n\log(a_n)$ is less than $-1$, so the series of $a_n\log(a_n)$ diverges to $-\infty$.
