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Question: Let $n > 0$. How can I find a function $f:\mathbb{N}\rightarrow\mathbb{R}^+$ such that $$ \lim_{n\to\infty} \frac{f(n)^2}{n} \log \left(\frac{f(n)}{n}\right) = L $$ with $0<L<\infty$?

Background: The term above appears in my research on subexponential bounds for binary words containing a limited number of ones. I have been able to elimate all other terms, but I am stuck with this one.

What I tried so far: I applied L'Hôpital's rule to get $$ \lim_{n\to\infty} \frac{\log\left(\frac{f(n)}{n}\right)}{\frac{-n}{f(n)^2}} = \lim_{n\to\infty} \frac{\frac{f'(n)}{f(n)}-\frac{1}{n}}{\frac{1}{f(n)^2}-\frac{2nf'(n)}{f(n)^3}} $$

which got rid of the $\log()$. Since the limit should be finite, it seems to me that $\lim_{n\to\infty} \frac{f(n)}{\sqrt{n}} < \infty$, but I haven't been able to come up with an $f(n)$ that doesn't lead to $L=0$.

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  • $\begingroup$ By $f(n)^2$ do you mean $(f(n))^2$ or $f(n^2)$ ? $\endgroup$ Apr 6, 2022 at 13:48
  • $\begingroup$ Also, does $f$ have to be a function with integer outputs? Or it can have real, non-integer outputs? Lots to clarify in the question in my opinion... $\endgroup$ Apr 6, 2022 at 14:01
  • $\begingroup$ @AdamRubinson with $f(n)^2$, I mean $f(n)f(n)=(f(n))^2$ $\endgroup$
    – mto_19
    Apr 6, 2022 at 15:02

2 Answers 2

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Take $f(n) = n+1$. Then \begin{align*} \lim\limits_{n\to\infty} \frac{f(n)^2}{n}\log\bigg(\frac{f(n)}{n}\bigg) = \lim\limits_{n\to\infty} \bigg(n + 2 + \frac 1n\bigg)\log\bigg(1 + \frac 1n\bigg) = \lim\limits_{n\to\infty} \frac{n + 2 + \frac 1n}{n}\cdot n\log\bigg(1 + \frac 1n\bigg) = 1. \end{align*} You can get any other number $L>0$ by taking $f(n) = n+L$.

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I'll assume that $f$ is a function which can output real values.

For each $n\in\mathbb{N},$ define $g_n:[n,\infty)\to\mathbb{R};\ g_n(x) = \frac{x^2}{n} \log\left(\frac{x}{n}\right).$

One can readily check that $g_n(x)$ is a continuous, increasing function in $x$, with range $[0,\infty).$

Therefore (by IVT), for each $n,\ \exists\ x_n\in [n,\infty)\ $ such that $ g_n(x_n)= \frac{{x_n}^2}{n} \log\left(\frac{x_n}{n}\right) = L.$

Let $f(n) = x_n$ for each $n\in\mathbb{N},$ and we have completed our construction of $f.$

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