pove that: $$\sum_{k=n}^{2n-3}\dfrac{|\sin{k}|}{k}<\dfrac{1}{\sqrt{2}},n\ge 3$$
This problem is my frend creat it today,Thank you someone can prove it.Thank you
my idea,long ago I have prove it $$\sum_{k=n}^{3n-1}\dfrac{|\sin{k}|}{k}>\dfrac{1}{6}$$ I use this $$\min{\{|\sin{x}|,|\sin{(x+1)}|}\}>\dfrac{1}{3},x\in R$$
If you bound $|\sin(k)|$ by $1$, then
$$ \begin{align} \sum_{k=n}^{2n-3}\frac{|\sin k|}{k}& < \sum_{k=n}^{2n-3}\frac{1}{k}\\ &<\int_{n-1}^{2n-3}\frac{1}{x}\,dx\\ &=\ln\left(\frac{2n-3}{n-1}\right)\\ &=\ln\left(2-\frac{1}{n-1}\right)\\ &<\ln\left(2\right)\\ \end{align} $$
And $\ln(2)<\frac{1}{\sqrt{2}}$.
Here is an improved bound. For $n\geq6$, at most one-half plus one of the values of $|\sin k|$ for $k$ in $\{k,k+1,\ldots,2k-6\}$ are greater than $1/\sqrt{2}$. These "large" values are spread out over $\{k,k+1,\ldots,2k-6\}$ but for the purposes of establishing an upper bound, we can put them all up front:
$$ \begin{align} \sum_{k=n}^{2n-6}\frac{|\sin k|}{k}& < \sum_{k=n}^{\lceil3n/2\rceil-2}\frac{1}{k}+\sum_{k=\lceil3n/2\rceil-1}^{2n-6}\frac{1}{k\sqrt{2}}\\ &<\int_{n-1}^{\lceil3n/2\rceil-2}\frac{1}{x}\,dx+\int_{\lceil3n/2\rceil-2}^{2n-6}\frac{1}{x\sqrt{2}}\,dx\\ &=\ln\left(\frac{\lceil3n/2\rceil-2}{n-1}\right)+\frac{1}{\sqrt{2}}\ln\left(\frac{2n-6}{\lceil3n/2\rceil-2}\right)\\ &\le\ln\left(3/2\right)+\frac{1}{\sqrt{2}}\ln\left(4/3\right)\approx0.6089\\ \end{align} $$
You could tweak this to improve it. In the way that this cut in half, you could cut in thirds, etc. And if you want to make this more like the original question, you can bound the $2n-5$th, $2n-4$th, $2n-3$rd, and $2n-2$nd, terms by $1/(2n-5)+1/(2n-4)+1/(2n-3)+1/(2n-2)$. And always look directly at the sum for small $n$.
