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Suppose we have that, for fixed $m$, $$ a_{n,m}

Do we then have that $$ \limsup_{n\to\infty}\frac{1}{n}\ln a_{n,m}

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    Fourth installment of the same question, the first having already been answered with everything there is to say. Maybe this is enough, no? If you still do not get why $<$ is wrong and $\leqslant$ is ok, receiving again the same explanations and examples will not help. Please stop abusing the site. – 2017-01-28

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Let $b_{n,m} := \exp(n), a_{n,m} := \exp(n - 1)$ for all $m$. Then $a_{n,m}

But $\limsup_{n\to\infty} \frac 1n\ln a_{n,m} = \limsup_{n\to\infty} \frac 1n\ln b_{n,m}$. So it is only "$\le$", not "$<$".