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The conjecture is true. For $t\in[0,1]$, let $N(t)$ be the number of $k\in\{1,2,\dotsc,n\}$ satisfying $\{n/k\}<t$. On the one hand, $$\sum_{k=1}^n f\left(\frac{n}{k}\right)=\int_0^{1/2}\bigl(N(t+1/2) …

See Erick Wong's response here. In particular, Kevin Ford proved (in more precise form) that $$ f(n) = \frac{n}{\log n} \exp\left(O(\log \log \log n)^2\right),$$ whence $f(n)/n$ tends to zero. The sam …

This question was considered by Wintner (1945). On pages 16-18 of the linked document, he observes that $\sum_n 2^{\omega(n)}|x_n|<\infty$ implies the absolute convergence of the second display (here …

No, it doesn't. It is easy to see that for any positive integer $m$ we have $$ \left|\frac{\sin(m)}{m}\right| + \left|\frac{\sin(m+1)}{m+1}\right| > \frac{1}{6m}.$$ Summing this up over all even posi …

The sequence $(x_n)$ does not have a limit. Let us assume, for a contradiction, that the limit exists. The limit cannot be nonzero or $\pm\infty$, because then $|m+1-nx_n|\to\infty$ by the triangle in …

For the coefficients $2^n$ the equidistribution theorem fails. In fact it is easy to exhibit an irrational $a$ such that the sequence $(2^na)_{n=0}^\infty$ is not even dense in $(0,1)$ modulo $1$. For …

The limit equals $m/2$ when $m>1$ or $k\geq 1$, and it equals $m$ when $m=1$ and $k=0$. I provide the proof for $m>1$ below, the other cases being very similar. Assume that $m>1$. The integral equals …

We have $f(\gamma,\beta)=0$ for every $\gamma>0$ and $\beta\in\mathbb{R}$. Indeed, $1-\gamma^{1/x}$ is asymptotically $(\log\gamma)/x$, and $(\log x)^\beta/x$ tends to zero for any $\beta\in\mathbb{R} …

The inequality you want to prove is false. For example, $$f_5(1/2,1)=0.6096\dots,$$ while $$\lim_{k\to\infty}f_k(1/2,1)=0.6398\dots$$