let $$f(x,y)=\dfrac{\arcsin{\dfrac{x}{y}}}{x}$$
show that $$\sum_{i=1}^{n}f(i,n)<\dfrac{3}{2},n>3$$
My try: since $$\sum_{i=1}^{n}f(i,n)=\sum_{i=1}^{n}\dfrac{\arcsin{\dfrac{i}{n}}}{i}$$ so I can find this limit $$\lim_{n\to infty}\sum_{i=1}^{n}f(i,n)=\lim_{n\to\infty}\dfrac{1}{n}\sum_{i=1}^{n}\dfrac{\arcsin{\dfrac{i}{n}}}{\dfrac{i}{n}}=\int_{0}^{1}\dfrac{\arcsin{x}}{x}dx$$
But for this inequality,I can't prove it. Thank you
I think this is deceptively easy. The sum in question $S(n)$ is a right-hand sum of an increasing function on $[0,1]$. Hence we have $S(n)>S(n+1)$ for each $n$. We need only check that $S(3)<3/2$. Actually I think we have $S(2)<3/2$.
I hope and wish that this will help you for the last part of your problem.
Concerning the last integral of the post, it can be established that $$\int\dfrac{\arcsin{x}}{x}dx=\sin ^{-1}(x) \log \left(1+e^{-2 i \cos ^{-1}(x)}\right)-\frac{1}{2} i \left(\sin ^{-1}(x)^2+\text{Li}_2\left(e^{2 i \sin ^{-1}(x)}\right)\right)$$ So $$\int_{0}^{1}\dfrac{\arcsin{x}}{x}dx=\frac{1}{2} \pi \log (2)=1.08879 $$ Another way to approach the problem is to expand $\frac{\sin ^{-1}(x)}{x}$ as a Taylor series built around $x=0$. This leads to an infinite sum of weighted binomial coefficients for the same value.
If the Taylor series is truncated to the first terms, this leads to $$1+\frac{x^2}{6}+\frac{3 x^4}{40}+\frac{5 x^6}{112}+\frac{35 x^8}{1152}+\frac{63 x^{10}}{2816}+\frac{231 x^{12}}{13312}+O\left(x^{13}\right)$$ Integration between $0$ and $1$ leads to a value close to $1.08368$; doubling the number of terms for the Taylor expansion leads to a value close to $1.08678$.
