Let $(a^n)_{n \in \mathbb{N}} \subset \mathbb{R}^k$ be a sequence of vectors with non-negative components and with $\sum_{i=1}^k a^n_i=1$ for all $n\in\mathbb{N}$. Then we have for each $n\in\mathbb{N}$ by the log-sum inequality:

$\sum_{i=1}^k a_i^{n+1} \log\dfrac{a_i^{n+1}}{a_i^n} \geq \left(\sum_{i=1}^k a_i^{n+1}\right) \log\dfrac{\sum_{i=1}^k a_i^{n+1}}{\sum_{i=1}^k a_i^{n}}=0,$

where the equality holds if and only if $a_i^{n+1}/a_i^n$ is constant for any $i=1,...,k$.

Now assume that

$\lim\limits_{n\rightarrow \infty}\sum_{i=1}^k a_i^{n+1} \log\dfrac{a_i^{n+1}}{a_i^n} = 0.$

Can we conclude that $\lim\limits_{n\rightarrow \infty} \dfrac{a_i^{n+1}}{a_i^n} = 1$? Intuitively yes, but how can this be proven rigorously?

Thanks in advance!

  • $\begingroup$ I find your notation $a_i^n$ confusing and keep wanting to see the $n$ as an exponent rather than an index. I would write this as $a_{i, n}$ or $a_{n, i}$. $\endgroup$ Dec 17, 2018 at 18:18

1 Answer 1


Not necessarily. For instance, put $k=2$ and for each $n\in\Bbb N$ put $a^n_1=\frac 1{2^n}$ and $a^n_2=1-\frac 1{2^n}$. Then

$$\lim_{n\to\infty}\sum_{i=1}^k a_i^{n+1} \log\dfrac{a_i^{n+1}}{a_i^n}= \lim_{n\to\infty}\frac 1{2^{n+1}}\log\dfrac{\frac 1{2^{n+1}}}{\frac 1{2^{n}}}+ \left(1-\frac 1{2^{n+1}}\right)\log\dfrac{1-\frac 1{2^{n+1}}}{1-\frac 1{2^{n}}}=0,$$

but $$\lim\limits_{n\rightarrow \infty} \dfrac{a_1^{n+1}}{a_1^n} = \frac 12.$$


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