$$\sum\nolimits\arccos\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}, n\in\mathbb{N}^{*}$$

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!


As a thumb rule, if the terms of a convergent series are really ugly and you know in advance that such a series has a nice closed form, then the given series is a telescopic one, very likely. For any $n\geq 1$ we have

$$\small\cos\left(\arccos\frac{n}{2n+1}-\arccos\frac{n+1}{2n+3}\right)\\ =\small \frac{n(n+1)}{(2n+1)(2n+3)}+\small\sqrt{\small\left(1-\frac{n^2}{(2n+1)^2}\right)\left(1-\frac{(n+1)^2}{(2n+3)^2}\right)}$$ hence $$ \sum_{n=1}^N \arccos\frac{n(n+1)+\sqrt{(n+1)(3n+1)(n+2)(3n+4)}}{(2n+1)(2n+3)} $$ is simply given by $\arccos\frac{1}{3}-\arccos\frac{N+1}{2N+3}$. I guess you can compute the limit as $N\to +\infty$.
It is given by $\arctan(2\sqrt{2})-\frac{\pi}{3}=\frac{\pi}{6}-\arctan\left(\frac{1}{2\sqrt 2}\right)\approx 0.183761866\approx \frac{43}{234}$.

  • 1
    $\begingroup$ You forgot to mention "By creative telescoping", which is what I find to be your trademark. $\endgroup$ – Simply Beautiful Art Oct 23 '17 at 0:51

To know the answer, first you must know some results:

the propose (1) is sum of angle of cosine:

$$\arccos\left(x_1\right)-\arccos\left(x_2\right)=\arccos\left(x_1\cdot x_2+\sqrt{\left(1-x_1^2\right)\left(1-x_2^2\right)}\ \right)\quad, x_1<x_2$$


$$ \arccos\left(x_1\right)-\arccos\left(x_2\right)=\arccos\left(x_1\cdot x_2+\sqrt{\left(1-x_1\right)\left(1+x_1\right)\left(1-x_2\right)\left(1+x_2\right)}\right)\tag{1}$$

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.



hence using limits I have:

$$\lim_{k\to\infty}S(k)=\lim_{k\to\infty}\left(\arccos\left(\frac{1}{3}\right)-\arccos\left(\frac{k+1}{2k+1}\right)\ \right)$$

$$\lim_{k\to\infty}S(k)=\arccos\left(\frac{1}{3}\right)-\lim_{k\to\infty}\left(\arccos\left(\frac{k+1}{2k+1}\right)\ \right)$$

$$\lim_{k\to\infty}S(k)=\arccos\left(\frac{1}{3}\right)-\arccos\left(\lim_{k\to\infty}\frac{k+1}{2k+1}\right)\ $$

$$\lim_{k\to\infty}S(k)= \arccos\left(\frac{1}{3}\right)-\arccos\left(\frac{1}{2}\right)$$


$$\sum_{n=1}^{\infty }\arccos\left(\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}\ \right)=\arccos\left(\frac{1}{3}\right)-\arccos\left(\frac{1}{2}\right)$$

that which is the same:

$$\sum_{n=1}^{\infty }\arccos\left(\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}\ \right)=\arccos\left(\frac{1+2\sqrt{6}}{6}\right)$$


This is the best I could do:

first I substituted $n$ with $\dfrac{1}{x}$ to try a MacLaurin expansion and I got

$$\arccos\frac{\sqrt{8 x^4+42 x^3+67 x^2+42 x+9}+x+1}{3 x^2+8 x+4}$$

then with the help of Mathematica I found that this is

$\dfrac{x^2}{2 \sqrt{3}}+O(x^3)$

resetting $x=\dfrac{1}{n}$ I got

$\dfrac{1}{(2\sqrt{3} )n^2}$ which converges

Actually making a table of the values of $a_n$ and $\dfrac{1}{(2\sqrt{3} )n^2}$ you can see that they are very close.

No clue about founding the sum. Just an approximate value $0.18$


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