How to find $$\sum_{i=1}^{n} \sin^{-1}\left(\frac{1}{i(i+1)}\right)?$$ I am trying this question by writing $1$ as $(i+1)-i$. So, the above summation will become $$\sum_{i=1}^{n} \sin^{-1}\left(\frac{(i+1)-i}{i(i+1)}\right).$$ So, the above summation will become $$\sum_{i=1}^{n} \sin^{-1}\left(\frac{1}{i}-\frac{1}{i+1}\right).$$ Now, putting the values of $i=1,2,3,\ldots$, we get the summation to be $$\sin^{-1}\frac{1}{2}+\sin^{-1}\frac{1}{6}+\sin^{-1}\frac{1}{12}+\ldots+\sin^{-1}\frac{1}{n(n+1)}.$$
But after this step, I can't find the summation. I mainly need the formula of summation of the above inverse trigonometric series up to $n$ terms. So, that I can find the summation for any values of $n$.
$$S_n=\sum _{i=1}^n \sin ^{-1}\left(\frac{1}{i (i+1)}\right)=\frac \pi 6+\sum _{i=2}^n \sin ^{-1}\left(\frac{1}{i (i+1)}\right)$$ Using the expansion of the arcsine and Taylor series
$$\sin ^{-1}\left(\frac{1}{i (i+1)}\right)=\sum_{m=2}^\infty (-1)^m\,\frac {a_m}{i^m}$$ where the first coefficients are $$\left\{1,1,1,1,\frac{7}{6},\frac{3}{2},2,\frac{8}{3},\frac{143}{4 0},\frac{39}{8},\frac{163}{24},\frac{77}{8},\frac{1545}{112}, \frac{4783}{240},\cdots\right\}$$
Using generalized harmonic numbers $$\sum _{i=2}^n \frac 1{i^m}=H_n^{(m)}-1$$
$$S_n=\frac \pi 6+\sum_{m=2}^\infty (-1)^m\,{a_m}\,\big(H_n^{(m)}-1\big)$$ We know that $$H_n^{(m)}=\zeta (m)+n^{-m} \left(-\frac{n}{m-1}+\frac{1}{2}-\frac{m}{12 n}+O\left(\frac{1}{n^3}\right)\right)$$
Using the terms given above, we end with $$S_n=C_{15}-\frac{1}{n}+\frac{1}{n^2}-\frac{1}{ n^2}-\frac{1}{n^3}+\frac{1}{n^4}+O\left(\frac{1}{n^5}\right)$$ where the constant is a linear combination of zeta functions.
Converted to decimals $$C_{15}=\color{red}{1.02425}203$$ $$C_{100}=\color{red}{1.02450817968}912$$
The conergence is not very fast because $$\frac{\sin ^{-1}\left(\frac{1}{(i+1) (i+2)}\right)}{\sin ^{-1}\left(\frac{1}{i (i+1)}\right)}=1-\frac{2}{i}+\frac{4}{i^2}+O\left(\frac{1}{i^3}\right)$$
Inverse symbolic calculators do not find anything looking like $$1.024508179688909499053001\cdots$$
Edit
Just to stay with a sum of positive terms, we could also write $$S_n=\frac \pi 6+\frac 1 {\sqrt \pi}\sum_{m=2}^n \frac{\Gamma \left(m+\frac{1}{2}\right)}{(2 m+1) \Gamma (m+1)}\frac 1 {\big(i(i+1)\big)^{2m+1}}$$ Even if not very nice, we know what is
$$T_m=\sum_{i=1}^n\frac 1 {\big(i(i+1)\big)^{2m+1}}$$ (polynomial in $n$ and polygamma functions).
Under this form, the convergence is quite faster since adding five terms only, we already have $$\color{red}{1.02450817968}755$$