If $\sum_{k\geq1}ka_k = 1$, then $\sum_{k\geq1}-a_k\log(a_k) < \infty$? Let $(a_k)_{k\geq1}$ be a sequence of non-negative numbers such that
$$ \sum_{k\geq1}ka_k < +\infty. $$
Is is true that
$$ \sum_{k\geq1}-a_k\log(a_k) < +\infty ?\qquad (1)$$
I think that maybe doing a bound like $-a_k\log(a_k) \leq a_k^{1/2}$ it is possible to prove (1), but I couldn't even prove this bound (altough wolfram alpha says that it is true).

Notice $$\frac{d}{dx} -x\log x = -(1+\log x) >0$$ for $x < e^{-1}$. 
Then $$\sum_{n =1}^\infty -a_n\log a_n = \sum_{n: a_n \leq e^{-n}} -a_n \log a_n + \sum_{n: a_n > e^{-n}} -a_n \log a_n$$ $$ \leq \sum_{n: a_n \leq e^{-n}} -a_n \log a_n + \sum_{n: a_n > e^{-n}} na_n \leq \sum_{n = 1}^\infty ne^{-n} + na_n < \infty $$
