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

Question

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).

Answer 1

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

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

Answer 2

Put $ka_k=\frac{kb_k}{2^k}$

The function :

$$f(x)=-x\ln(x)$$ is concave and as the condition is finite because we have $\sum_{k=1}^{\infty}ka_k=\sum_{k=1}^{\infty}\frac{kb_k}{2^k}< +\infty$ we can apply the infinite discrete form of Jensen's inequality we got :

$$\sum_{k=1}^{\infty}-\frac{b_k}{2^k}\log(\frac{b_k}{2^k})=\sum_{k=1}^{\infty}-\frac{b_k}{2^k}\log(b_k)+\frac{kb_k}{2^k}\log(2)\leq -\Big(\sum_{k=1}^{\infty}(\frac{b_k}{2^k})\Big)\log(\sum_{k=1}^{\infty}(\frac{b_k}{2^k}))+\frac{kb_k}{2^k}\log(2)<+\infty$$