<s>[INST] Show that series is convergent and find its sum $$\sum\nolimits\arccos\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}, n\in\mathbb{N}^{*}$$
I need help proving that this series is convergent and calculating its sum.
What I've done so far:
$$\lim_{n\to\infty}\arccos\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}=0$$
The result shows that the series may be convergent, but I don't know how to continue. WolframAlpha shows that the series is convergent (by the comparison test), but I have no idea to what other convergent series could I compare it.
Thank you!
 [/INST] To know the answer, first you must know some results:
the propose (1) is sum of angle of cosine:
$$\arccos\left(x_1\right)-\arccos\left(x_2\right)=\arccos\left(x_1\cdot x_2+\sqrt{\left(1-x_1^2\right)\left(1-x_2^2\right)}\ \right)\quad, x_1<x_2$$
hence
$$ \arccos\left(x_1\right)-\arccos\left(x_2\right)=\arccos\left(x_1\cdot x_2+\sqrt{\left(1-x_1\right)\left(1+x_1\right)\left(1-x_2\right)\left(1+x_2\right)}\right)\tag{1}$$
So compare the expression (1) with expression (2):
$$\arccos\left(\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}\ \right)\tag{2}$$
hence can be easily deduced:
$$x_1=\frac{n}{2n+1}\ \ ,\ \ \ x_2=\frac{n+1}{2n+3}\tag{3}$$
where $x_1<x_2$.
so using (3) in (1) obtain:
$$S(k)=\sum_{n=1}^k\arccos\left(\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}\ \right) $$
$$ S(k)=\sum_{n=1}^k\left(\arccos\left(\frac{n}{2n+1}\right)-\arccos\left(\frac{n+1}{2n+3}\right)\right)$$
$$ S(k)=\sum _{n=1}^{k }\arccos\left(\frac{n}{2n+1}\right)-\sum _{n=1}^{k }\arccos\left(\frac{n+1}{2n+3}\right)$$
it is easy to see that this is a telescoping series.
So.
$$S(k)=\arccos\left(\frac{1}{3}\right)-\arccos\left(\frac{k+1}{2k+1}\right)$$
hence using limits I have:
$$\lim_{k\to\infty}S(k)=\lim_{k\to\infty}\left(\arccos\left(\frac{1}{3}\right)-\arccos\left(\frac{k+1}{2k+1}\right)\ \right)$$
$$\lim_{k\to\infty}S(k)=\arccos\left(\frac{1}{3}\right)-\lim_{k\to\infty}\left(\arccos\left(\frac{k+1}{2k+1}\right)\ \right)$$
$$\lim_{k\to\infty}S(k)=\arccos\left(\frac{1}{3}\right)-\arccos\left(\lim_{k\to\infty}\frac{k+1}{2k+1}\right)\ $$
$$\lim_{k\to\infty}S(k)= \arccos\left(\frac{1}{3}\right)-\arccos\left(\frac{1}{2}\right)$$
Finally:
$$\sum_{n=1}^{\infty }\arccos\left(\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}\ \right)=\arccos\left(\frac{1}{3}\right)-\arccos\left(\frac{1}{2}\right)$$
that which is the same:
$$\sum_{n=1}^{\infty }\arccos\left(\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}\ \right)=\arccos\left(\frac{1+2\sqrt{6}}{6}\right)$$
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