How can I get the exact value of this infinite series? I want to compute the exact value of this infinite series 
$$\sum_{n=2}^\infty\arcsin{\left(\dfrac{2}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n-1})}\right)}$$
By comparison test, we can get the series is convengence.
I tried to find some hints from  the exact value of $\displaystyle\sum_{n=2}^\infty\arcsin{\left(\dfrac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n^2-1}}\right)}$
,but the split phase method maybe difficult to solve this  question. 
I am not sure whether it has  a closed form.But if not so,how can I evaluate the sum ?
 A: The sum to 1000 terms,
according to Wolfy,
is
1.55713
or $0.481892\pi$.
Would be amusing if the sum were
$\pi/2$.
$\sum_{n=2}^\infty\arcsin{\left(\dfrac{2}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n-1})}\right)}
$
More seriously,
I will try
$\arcsin(x)
=\arctan(\frac{x}{\sqrt{1-x^2}})
$
to see if
$\arctan(u)+\arctan(v)
=\arctan(\frac{u+v}{1-uv})
$
can be used.
If
$x
=\dfrac{2}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n-1})}
=\dfrac{2}{f(n)}
$,
$\begin{array}\\
\frac{x}{\sqrt{1-x^2}}
&=\frac{\frac{2}{f(n)}}{\sqrt{1-(\frac{2}{f(n)})^2}}\\
&=\frac{2}{\sqrt{f^2(n)-2}}\\
&=\frac{2}{\sqrt{n(n+1)(2n-1+2\sqrt{n(n-1)}-2)}}\\
&=\frac{2}{\sqrt{n(n+1)(2n-3+2\sqrt{n(n-1)})}}\\
\end{array}
$
For $n=2$,
this is
$\frac{2}{\sqrt{6(1+2\sqrt{2})}}
$.
For $n=3$,
this is
$\frac{2}{\sqrt{12(3+2\sqrt{6})}}
$.
Combining these,
to get the sum up to 3,
we get
$\begin{array}\\
\dfrac{\frac{2}{\sqrt{6(1+2\sqrt{2})}}+\frac{2}{\sqrt{12(3+2\sqrt{6})}}}{1-\frac{2}{\sqrt{6(1+2\sqrt{2})}}\frac{2}{\sqrt{12(3+2\sqrt{6})}}}
&=2\dfrac{\sqrt{6(1+2\sqrt{2})}+\sqrt{12(3+2\sqrt{6})}}{\sqrt{6(1+2\sqrt{2})}\sqrt{12(3+2\sqrt{6})}-4}\\
&=2\dfrac{\sqrt{6(1+2\sqrt{2})}+\sqrt{12(3+2\sqrt{6})}}{6\sqrt{2(1+2\sqrt{2}) (3+2\sqrt{6})}-4}\\
&=2\dfrac{\sqrt{6(1+2\sqrt{2})}+\sqrt{12(3+2\sqrt{6})}}{6\sqrt{2(3+6\sqrt{2}+2\sqrt{6}+4\sqrt{12})}-4}\\
&=\dfrac{\sqrt{6(1+2\sqrt{2})}+\sqrt{12(3+2\sqrt{6})}}{3\sqrt{2(3+6\sqrt{2}+2\sqrt{6}+8\sqrt{3})}-2}\\
\end{array}
$
This doesn't look like
anything I would like
to meet in a dark alley.
So I'll give it up here.
A: This is not an answer but it is too long for a comment.
I do not think about any possible closed form.
Considering $$u_n=\arcsin{\left(\dfrac{2}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n-1})}\right)}$$ Taylor expansion for large values of $n$ gives $$u_n\simeq \left(\frac{1}{n}\right)^{3/2}-\frac{1}{4} \left(\frac{1}{n}\right)^{5/2}+\frac{3}{8}
   \left(\frac{1}{n}\right)^{7/2}-\frac{7}{192}
   \left(\frac{1}{n}\right)^{9/2}+\frac{17}{128}
   \left(\frac{1}{n}\right)^{11/2}+\cdots $$ So, approximating the summation $$\sum_{n=2}^\infty u_n \simeq \frac{1}{384} \left(384 \zeta \left(\frac{3}{2}\right)-96 \zeta
   \left(\frac{5}{2}\right)+144 \zeta \left(\frac{7}{2}\right)-14 \zeta
   \left(\frac{9}{2}\right)+51 \zeta \left(\frac{11}{2}\right)-469\right)$$ At this level of truncation, the result is $\approx 1.57588 $. Increasing the number of terms in the expansions, we arrive to the value already mentioned in comments and answers $\approx1.57713$.
