Show that $$\sum_{k=1}^n{\mspace{-2mu}\frac{\left\lvert\sin{k}\right\rvert}{k}}\sim\frac{2}{\pi}\mspace{-1.5mu}\sum_{k=1}^n{\mspace{-2mu}\frac{1}{\mspace{-1mu}k}}$$ as $n\to\infty$.
Alternatively, since $\displaystyle\frac{1}{\ln{x}}\mspace{-1.5mu}\int_0^x{\mspace{-2mu}\frac{\left\lvert\sin{t}\right\rvert}{t}\operatorname{d}\!t}$ converges to $\dfrac{2}{\pi}$ as $x\to{+\infty}$ and $\displaystyle\lim_{n\to\infty}{\frac{{\it{H}}_n}{\ln{n}}}=1$, how can we prove that $$\sum_{{1}\leq{n}\leq{\left\lfloor{x}\right\rfloor}}{\mspace{-2mu}\frac{\left\lvert\sin{n}\right\rvert}{n}}\sim\int_0^x{\mspace{-2mu}\frac{\left\lvert\sin{t}\right\rvert}{t}\operatorname*{d}\!t}$$ as $x\to{+\infty}$?
- It may seem like just one simple application of the so-called Euler–Maclaurin formula. Nonetheless, $\dfrac{\left\lvert\sin{u}\right\rvert}{u}$ is not always continuously differentiable on $\left[0, x\right]$, hence the use of the Euler–Maclaurin summation formula shall be inadmissible in fact.
- So does the Stolz–Cesàro theorem.
Some "similar" problems can be seen in many posts such as How can we prove that …, How to prove the convergence of the series? and How to find the limit….
Best Answer
If we compute the integrals over multiples of $\pi$, the result is $$I_k=\int_{2 \pi k}^{2 \pi (k+1)} \frac{| \sin (t)| }{t} \, dt=2\, \text{Si}(2(k+1) )-\text{Si}(2 k \pi )-\text{Si}(2 (k+1) \pi )$$
Using the asymptotics of the trigonometric integrals, $$I_k=\frac{2}{\pi k}-\frac{1}{\pi k^2}+\frac{3 \pi ^2-4}{4 \pi ^3 k^3}+\frac{12-5 \pi ^2}{8 \pi ^3 k^4}+\frac{48-36 \pi ^2+9 \pi ^4}{16 \pi ^5 k^5}+O\left(\frac{1}{k^6}\right)$$ The summation followed by the expansion of the generated generalized harmonic numbers gives
$$\sum_{k=1}^n I_k=C+\frac{2 }{\pi }\log (n)+\frac{2}{\pi n}-\frac{25 \pi ^2-12}{24 \pi ^3 n^2}+O\left(\frac{1}{n^3}\right)$$ where, for this level of expansion, $C \sim \frac 1 {20}$.