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Consider a sequence of equations $$ \exp\left( \frac{k}{1+\varepsilon k}\right) = \frac{\text{e}^{-k}m_k}{n_k+\text{e}^{-2k}m_k}, \;\;\; k=1,2,\ldots $$ on numbers $m_k, n_k >0$. Here $\varepsilon > 0$ is a parameter. Is it possible to find solution $(m_1,m_2,\ldots)$, $(n_1,n_2,\ldots)$ such that $$ 0<\inf_{k} n_k \leqslant\sup_{k} n_k < \infty \\ \sup_{k} m_{k} < \infty ? $$

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Rewrite the above equation as $\exp{\left ( \frac{k (2 + \epsilon k)}{1 + \epsilon k} \right )} + \exp{(-2 k)} = -\frac{n_k}{m_k} $, see if that helps. – Ron Gordon Jan 11 '13 at 19:04
@rlgordonma good advice, $\frac{m_k}{n_k} = \exp\left( \frac{k(2+\varepsilon k)}{1+\varepsilon k} \right) \left(1 - \exp\left(\frac{-\varepsilon k^2}{1+\varepsilon k}\right) \right)^{-1}$, LHS is bounded whereas RHS not, thank you! – Nimza Jan 11 '13 at 22:00
You're welcome. – Ron Gordon Jan 11 '13 at 23:04

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