# Evaluating a given limit

Let $$\nu(x) = \frac{1}{2} \log{2\pi x} - x + \int\limits_{1}^{x} \frac{\lfloor{t\rfloor}}{t} \ dt$$ then what is $\displaystyle \lim_{x \to \infty} x \cdot \nu(x)$?

I really don't know to proceed. Tried something using the expansions of $\log$ and seeing how the integral behaves, but couldn't figure out.

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## 1 Answer

[The answer is given at the end.]

First let $n = \left\lfloor x \right\rfloor$. We have $\int_k^{k + 1} {\frac{{\left\lfloor t \right\rfloor }}{t}\,{\rm d}t} = k[\log (k + 1) - \log (k)]$, $k=1,\ldots,n-1$, and $\int_n^x {\frac{{\left\lfloor t \right\rfloor }}{t}\,{\rm d}t} = n[\log (x) - \log (n)]$. From this it follows straightforwardly that $$\nu(x) = \frac{1}{2}\log (2\pi x) - x - \log (n!) + n\log (x).$$ Next, let $\delta_x = x - n$. Thus, $x = n + \delta_x$. Substituting this for $x$ in the above expression, and using $$\log (M + a) = \log (M) + \frac{a}{M} - \frac{{a^2 }}{{2M^2 }} + O\bigg(\frac{1}{{M^3 }}\bigg),$$ as $M \to \infty$, leads to $$\nu (x) = \frac{1}{2}\log (2\pi n) - n - \log (n!) + n\log (n) + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg),$$ or $$\nu (x) = \log \Bigg[\frac{{(\sqrt {2\pi n} )e^{ - n} n^n }}{{n!}}\Bigg] + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg),$$ as $n \to \infty$. Then, from the well-known estimate $$n! = (\sqrt {2\pi n} )e^{ - n} n^n \bigg(1 + \frac{1}{{12n}} + O\bigg(\frac{1}{{n^2 }}\bigg)\bigg)$$ (and noting that $\log(1/\xi)=-\log(\xi)$), it follows that $$\nu (x) = - \log \bigg[1 + \frac{1}{{12n}} + O\bigg(\frac{1}{{n^2 }}\bigg)\bigg] + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg),$$ and in turn, $$\nu (x) = - \frac{1}{{12n}} + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg).$$ Finally, since $x = n + \delta_x$, we obtain $$x \nu(x) = - \frac{1}{{12}} + \frac{{\delta _x }}{{2}} - \frac{{\delta _x^2 }}{{2}} + O\bigg(\frac{1}{{n}}\bigg).$$ We conclude that $x \nu(x)$ does not have a limit as $x \to \infty$. Rather, since $\delta_x /2 - \delta_x^2 / 2$ is oscillating between $0$ and $1/8$, we have $$\mathop {\lim \inf }\limits_{x \to \infty } [x\nu (x)] = - \frac{1}{{12}} \;\;\; {\rm and} \;\;\; \mathop {\lim \sup }\limits_{x \to \infty } [x\nu (x)] = \frac{1}{{24}}.$$ Nevertheless, we have $$\mathop {\lim }\limits_{n \to \infty } n\nu (n) = - \frac{1}{{12}},$$ as $n \to \infty$ (integer).

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@Shai Covo: Yes, so then perhaps we can talk about the $\lim\inf$ or the $\lim\sup$ – anonymous Nov 3 '10 at 16:46
Specifically, I am going to show that, as $x \to \infty$, $x \nu(x) \approx -1/12 - \delta_x^2/2 + \delta_x/2$, where $\delta_x = x - \left\lfloor x \right\rfloor$, implying that the $\lim\inf$ is equal to $-1/12$ and the $\lim\sup$ to $1/24$. – Shai Covo Nov 3 '10 at 16:57
@Shai: Eagerly awaiting your answer. – anonymous Nov 3 '10 at 18:31
My calculations agree with Shai's. – Mike Spivey Nov 3 '10 at 19:50
I will post the solution shortly. – Shai Covo Nov 3 '10 at 20:00