I need to calculate the limit (if it exist) $$\lim_{n\to\infty}\left\{n\sum_{k=1}^n\frac{1}{k^5}\right\}$$ where $\{x\}$ denotes the fractional part of $x$ and $n\in\mathbb{N}$.
By definition of fractional part function, $$0\leq\left\{n\sum_{k=1}^n\frac{1}{k^5}\right\}<1$$ So we have $$0\leq\lim_{n\to\infty}\left\{n\sum_{k=1}^n\frac{1}{k^5}\right\}\leq1$$ Sorry but I don't have any ideas. Any help would be highly appreciated.
First, observe that $$\sum_{k=1}^n \frac{1}{k^5} = \zeta(5) - \sum_{k=n+1}^\infty \frac{1}{k^5}.$$ However, by an argument similar to the one used in the proof of the integral test, $$0 < \sum_{k=n+1}^\infty \frac{1}{k^5} < \int_n^\infty \frac{dx}{x^5} = \frac{1}{5 n^4}.$$ Therefore, if we let $a_n := n \sum_{k=1}^n \frac{1}{k^5}$, then $$n \zeta(5) - \frac{1}{5 n^3} < a_n < n \zeta(5).$$ However, at this point, if we try to get an approximation of $\{ a_n \}$ using the above approximation of $a_n$, then we run into the issue that the fractional part function is discontinuous at integers, so $\{ a_n \}$ and $\{ n \zeta(5) \}$ could be different by almost one when $a_n$ is close to an integer. To get around this, I will use the fairly standard trick of using the function $x \mapsto e^{2\pi i x}$ from $\mathbb{R}$ to the unit circle $S^1$ in the complex plane; this function is continuous and is closely related to the fractional part function. (Alternatively, if you're comfortable with the topological group $\mathbb{R} / \mathbb{Z}$, then you could instead use the projection map $\mathbb{R} \to \mathbb{R} / \mathbb{Z}$.)
Thus, from the above inequalities, it is straightforward to conclude that $a_{n+1} - a_n \to \zeta(5)$ as $n \to \infty$; as a result, $e^{2\pi i a_{n+1}} / e^{2\pi i a_n} \to e^{2\pi i \zeta(5)}$ as $n \to \infty$. However, since $1 < \zeta(5) < 2$, $\zeta(5)$ is not an integer, so $e^{2\pi i \zeta(5)} \ne 1$. That implies that the sequence $e^{2\pi i a_n}$ cannot be convergent.
On the other hand, $e^{2\pi i a_n} = e^{2\pi i \{ a_n \}}$. Therefore, if we had $\lim_{n\to \infty} \{ a_n \} = L$, we would have $e^{2\pi i a_n} \to e^{2\pi i L}$ as $n\to \infty$, giving a contradiction to the previous paragraph. We can thus conclude that $\{ a_n \}$ also is not a convergent sequence.
Assume a number $a$ is an accumulation point of the sequence $\{a_n\}.$ Then $0\le a\le 1.$ Observe that for any sequence $0<b_n\to 0$ the number $a$ is an accumulation point of $\{a_n-b_n\},$ when $0<a\le 1,$ and $1$ is an accumulation point of $\{a_n-b_n\},$ when $a=0.$ Thus $\{a_n-b_n\}$ is not convergent if $\{a_n\}$ admits at least two accumulation points.
We are going to apply the above to the arithmetic sequence $a_n=nx.$ It is well known that for an irrational number $x$ the sequence $\{nx\}$ is dense in $(0,1).$
Assume $\{x\}={p\over q},$ where $q\ge 2$ and $p$ and $q$ are relatively prime. The sequence $\{nx\}$ is $q$- periodic and its accumulation points are ${s\over q}$ for $0\le s<q.$
In both cases the sequence $\{nx\}$ has at least two accumulation points. So does $\{nx-b_n\}.$
Now we can apply the above to $x=\zeta(5)$ and $b_n=\displaystyle n\sum_{k=n+1}^\infty k^{-5},$ as $$n\zeta(5)-b_n=n\sum_{k=1}^nk^{-5}$$
Thus the sequence in question is divergent, provided that $\zeta(5)$ is not an integer.
