# 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).

• Who put -1 should write a comment about how to imporove my question! Commented May 11, 2020 at 15:56

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

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

• Niceee! I tried something similar but I couldn't make it work Commented May 11, 2020 at 17:18

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 :

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