I need to calculate the limit (if it exist) $$\lim_{n\to\infty}\left\{n!\sum_{k=1}^{n!}\frac{1}{k^{\frac{3}{2}}}\right\}$$ where $\{x\}$ denotes the fractional part of $x$, $n!$ is the factorial of $n$ and $n\in\mathbb{N}$.
We have $$ \sum_{k=1}^{n!}\frac{1}{k^{\frac{3}{2}}}=H^{(\frac{3}{2})}_{n!}$$ where $H^{(r)}_{n}$ is the generalized harmonic number. So we have$$\lim_{n\to\infty}\left\{n!\sum_{k=1}^{n!}\frac{1}{k^{\frac{3}{2}}}\right\}=\lim_{n\to\infty}\left\{n!H^{(\frac{3}{2})}_{n!}\right\}$$I have observed the behavior of the fractional part using Wolfram Alpha. The plot shows that $\left\{n!H^{(\frac{3}{2})}_{n!}\right\}$ oscillates between $y=0$ and $y=1$. So my gut feeling is that the sequence $\left\{n!H^{(\frac{3}{2})}_{n!}\right\}$ is not convergent. Also we have$$\sum_{k=1}^{n!}\frac{1}{k^{\frac{3}{2}}}=\sum_{k=1}^{\infty}\frac{1}{k^{\frac{3}{2}}}-\sum_{k=n!+1}^{\infty}\frac{1}{k^{\frac{3}{2}}}$$So we have $$n!\sum_{k=1}^{n!}\frac{1}{k^{\frac{3}{2}}}=n!\ \zeta\left(\frac{3}{2}\right)-n!\sum_{k=n!+1}^{\infty}\frac{1}{k^{\frac{3}{2}}}$$where $\zeta(s)$ denotes the Riemann zeta function. Now denote$$a_n=n!\sum_{k=1}^{n!}\frac{1}{k^{\frac{3}{2}}}$$So if the sequence $\left\{a_n\right\}$ is convergent, then $$\lim_{n\to\infty} \left\{a_n\right\}=l $$So we have $$\lim_{n\to\infty} \left\{a_{n+1}\right\}=l $$Now $$e^{2\pi i a_n}=e^{2\pi i \left\{a_n\right\}}$$Since $\lim_{n\to\infty} \left\{a_n\right\}=l$, so we have$$\lim_{n\to\infty} e^{2\pi i a_n}=e^{2\pi i l}$$ and$$\lim_{n\to\infty} e^{2\pi i a_{n+1}}=e^{2\pi i l}$$ So by dividing the above equations we get$$\lim_{n\to\infty} e^{2\pi i (a_{n+1}-a_{n})}=1$$Now we have$$a_{n+1}=(n+1)!\sum_{k=1}^{(n+1)!}\frac{1}{k^{\frac{3}{2}}} $$
Sorry but I don't have any nice ideas. Any help would be highly appreciated. Thank you.
Edit: We can express the sum in terms of the Hurwitz zeta function:$$\sum_{k=1}^{n!}\frac{1}{k^{3/2}}=\zeta(3/2)-\zeta(3/2,n!+1)$$ and the asymptotic behavior, as $n \to \infty$, is given by$$\sum_{k=1}^{n!}\frac{1}{k^{3/2}}= \zeta(3/2)-\frac{2}{\sqrt{n!+1}}-\frac12 \frac1{(n!+1)^{3/2}}+\mathcal{O}\left( \frac1{(n!)^{5/2}}\right)$$(see the link above). We would obtain the same result with the Euler-Maclaurin formula.
So we have $$a_n= n! \sum_{k=1}^{n!}\frac{1}{k^{3/2}}= n!\ \zeta(3/2)-\frac{2 n!}{\sqrt{n!+1}}-\frac{n!}{2} \frac1{(n!+1)^{3/2}}+\mathcal{O}\left( \frac1{(n!)^{3/2}}\right)$$Hence we obtain$$a_n=n!\ \zeta(3/2)-\frac{2\sqrt{n!} }{\sqrt{1+\frac{1}{n!}}}-\frac{1}{2 (n!)^{1/2}} \frac1{(1+\frac{1}{n!})^{3/2}}+\mathcal{O}\left( \frac1{(n!)^{3/2}}\right)$$So we obtain an asymptotic expansion of $a_n$ as $n\to\infty$, $$a_n\sim n!\ \zeta(3/2)-\frac{2\sqrt{n!} }{\sqrt{1+\frac{1}{n!}}}$$So we have $$\{a_n\}=\left\{ n!\ \zeta(3/2)-\frac{2\sqrt{n!} }{\sqrt{1+\frac{1}{n!}}}-\frac{1}{2 (n!)^{1/2}} \frac1{(1+\frac{1}{n!})^{3/2}}+\mathcal{O}\left( \frac1{(n!)^{3/2}}\right)\right\} $$