I've got to the problem of testing convergence using the integrals on $$ \sum _{n=1} ^{\infty} \arcsin \left( \frac{1}{\sqrt{x}} \right) $$ Our theory says:
Consider an integer $N$ and a non-negative function $f$ defined on the unbounded interval $[N, \infty)$, on which it is monotone decreasing. Then the infinite series $$ \sum_{n=N}^\infty f(n) $$ converges to a real number if and only if the improper integral $$ \int_N^\infty f(x)\,dx $$ is finite. In other words, if the integral diverges, then the series diverges as well.
But as anyone can clearly see, the function in the sum is not monotonely decreasing at all! What should I do with it? How to solve it?
Should I just try it with this? $$ \int _1 ^{\infty} \arcsin \left( \frac{1}{\sqrt{x}}\right)dx $$ If so, the help is also needed...
If we use the Integral Test the series will be diverge because $$\int_{1}^{\infty }\arcsin(\frac{1}{\sqrt{x}})dx=\sqrt{x-1}+x\arcsin(\frac{1}{\sqrt{x}}) $$ it is clear the value of integral will be infinity, so the series will be diverge