If $\sum_{k\geq1}ka_k = 1$, then $\sum_{k\geq1}-a_k\log(a_k) < \infty$?

Notice $$\frac{d}{dx} -x\log x = -(1+\log x) >0$$ for $x < e^{-1}$.

Then $$\sum_{n =1}^\infty -a_n\log a_n = \sum_{n: a_n \leq e^{-n}} -a_n \log a_n + \sum_{n: a_n > e^{-n}} -a_n \log a_n$$ $$ \leq \sum_{n: a_n \leq e^{-n}} -a_n \log a_n + \sum_{n: a_n > e^{-n}} na_n \leq \sum_{n = 1}^\infty ne^{-n} + na_n < \infty $$