# Minimization of $\sum \frac{1}{n_k}\ln n_k >1$ subject to $\sum \frac{1}{n_k}\simeq 1$

Looking at an algorithm for minimizing $\sum_{k=1}^{m} \frac{1}{n_k}\ln n_k > 1$ subject to $\sum_{k=1}^{m}\frac{1}{n_k} = 1$ in which $n_k$ are positive and in general non-sequential integers, I wondered about the more general problem of finding the minimum of $\sum_{k=1}^{m} \frac{1}{n_k}\ln n_k > 1$ subject to $\sum_{k=1}^{m}\frac{1}{n_k} \simeq 1$.

For example: $\frac{1}{2}+ \frac{1}{3}+\frac{1}{6} = 1$, and $\frac{1}{2}\ln 2+ \frac{1}{3}\ln 3 + \frac{1}{6}\ln 6 \simeq 1.014$.

We also have $\frac{1}{2}+\frac{1}{3}+\frac{1}{8} + \frac{1}{200}+\frac{1}{5000} \simeq .96$ with

$\frac{1}{2}\ln 2 +\frac{1}{3}\ln 3 + \frac{1}{8}\ln 8 + \frac{1}{200}\ln 200 +\frac{1}{5000}\ln 5000 \simeq 1.0009$.

Are there general ways of thinking about this? While I would think there are a finite number of solutions for $(\sum_{k=1}^{m} \frac{1}{n_k}\ln n_k – 1 )< \epsilon_1$ and $| \sum_{k=1}^{m}\frac{1}{n_k} – 1| \leq \epsilon_2$, and a countable number of solutions if m can be infinite, I don’t see any systematic way of finding solutions even in the finite case.

Thanks for any suggestions.

Edit: typo corrected–sense of inequality in $\epsilon_1$ expression was backward. Should conform to question in title and first paragraph above.

#### Solutions Collecting From Web of "Minimization of $\sum \frac{1}{n_k}\ln n_k >1$ subject to $\sum \frac{1}{n_k}\simeq 1$"

[EDITED: This answer isn’t relevant to the updated version of the question.] Given any (large) integer $B$, if you let $n_1=B$, …, $n_m = [eB]$, then
$$\sum_{k=1}^m \frac1{n_k} = 1 + O(1/B)$$
but
$$\sum_{k=1}^m \frac{\ln n_k}{n_k} > \ln B \sum_{k=1}^m \frac1{n_k} > \ln B + O(1).$$
So there are lots of solutions where the sum with ln is far larger than 1.