# 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 ?

• What is the origin of this series? That's an odd-looking term to just appear out of nowhere... Jul 11, 2015 at 3:18
• @vadim123: Oh,you are right .I reedit it from $n=2$ to $\infty$.
– user250236
Jul 11, 2015 at 3:21
• @user250236, perhaps you might give a hint about the surrounding problems and the content of the preceding chapter. Jul 11, 2015 at 3:36
• @user250236: Note, the linked solution is actually quite close. Multiply numerator and denominator by $(\sqrt{n}-\sqrt{n-1})$ to get just twice the linked argument. Change the 2 to 1 in the numerator and the problem is solved. Jul 11, 2015 at 4:27
• $\displaystyle\sum_{n=2}^\infty \arcsin\frac{\color{red}1}{\sqrt{n~(n+1)}\cdot\Big(\sqrt n+\sqrt{n-1}\Big)} ~=~ \frac\pi4$ Jul 11, 2015 at 12:59

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.

• The sum to a half-million terms is $0.5011\pi$. Jul 11, 2015 at 3:36
• This is not so much an answer as a comment.
– robjohn
Jul 11, 2015 at 3:39
• It was my answer that was a comment. But I have added some stuff to try to give it an aura of respectability. And I upvoted all the comments. Jul 11, 2015 at 4:16
• @martycohen: Since it is shown in this answer that the sum of the $\arcsin$ of the arguments divided by $2$ is $\pi/4$ and since $\arcsin(2x)\gt2\arcsin(x)$ for $x\in(0,\frac12]$, the current sum must be more than $\pi/2$
– robjohn
Jul 14, 2015 at 16:08

Not yet finding useful expressions to which to apply $$\arcsin(x)-\arcsin(y)=\arcsin\left(x\sqrt{1-y^2}-y\sqrt{1-x^2}\right)$$ for the given sum, I will explain how one can approximate the value of the sum.

First we can compute the series \begin{align} &\arcsin\left(\frac2{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n-1})}\right)\\[6pt] &\small=n^{-3/2}-\tfrac14n^{-5/2}+\tfrac38n^{-7/2}-\tfrac7{192}n^{-9/2}+\tfrac{17}{128}n^{-11/2}+\tfrac{23}{512}n^{-13/2}+\tfrac{1277}{15360} n^{-15/2}+\tfrac{227}{16384}n^{-17/2}\\ &\small+\tfrac{3669}{32768}n^{-19/2}-\tfrac{22005}{917504}n^{-21/2}+\tfrac{30157}{262144}n^{-23/2}-\tfrac{122493}{10485760}n^{-25/2}+\tfrac{3021997}{37748736}n^{-27/2}+\tfrac{325343}{16777216} n^{-29/2}\\ &\small+\tfrac{1877155}{33554432}n^{-31/2}+\tfrac{839806723}{35433480192}n^{-33/2}+\tfrac{4843811671}{75161927680}n^{-35/2}+\tfrac{62611265}{8589934592}n^{-37/2}+\tfrac{51911966591}{670014898176}n^{-39/2}\\ &\small+\tfrac{585844333}{137438953472}n^{-41/2}+\tfrac{19229654735}{274877906944}n^{-43/2}+\tfrac{1012455521821}{49478023249920}n^{-45/2}+O\left(n^{-47/2}\right) \end{align} Then apply the Euler-Maclaurin Sum Formula to get \begin{align} &\sum_{k=2}^n\arcsin\left(\frac2{\sqrt{k(k+1)}(\sqrt{k}+\sqrt{k-1})}\right)\\[6pt] &\small=C-2n^{-1/2}+\tfrac23n^{-3/2}-\tfrac25n^{-5/2}+\tfrac14n^{-7/2}-\tfrac5{36}n^{-9/2}+\tfrac{41}{704}n^{-11/2}-\tfrac{7}{416}n^{-13/2}+\tfrac{29}{1536}n^{-15/2}\\ &\small-\tfrac{209}{4352}n^{-17/2}+\tfrac{148975}{2179072}n^{-19/2}-\tfrac{9931}{172032}n^{-21/2}+\tfrac{758479}{15073280}n^{-23/2}-\tfrac{161979}{1638400}n^{-25/2}+\tfrac{6072637}{56623104}n^{-27/2}\\ &\small+\tfrac{4932997}{30408704}n^{-29/2}-\tfrac{13284961755}{40047214592}n^{-31/2}-\tfrac{10330566383}{9688842240}n^{-33/2}+\tfrac{12196256559}{5368709120}n^{-35/2}+\tfrac{1973513480505}{258234908672}n^{-37/2}\\ &\small-\tfrac{2054154874907}{111669149696}n^{-39/2}-\tfrac{61779979336283}{880468295680}n^{-41/2}+\tfrac{38613309561025709}{206845624975360}n^{-43/2}+O\left(n^{-45/2}\right) \end{align} Computing the sum of arcsines up to $n=1000$, then subtracting the non-constant part of the asymptotic expansion, we get the value $$C=1.577134230124889513864011795293057977459257178812501378053524$$

• I would find it useful to see how you did these computations. Somehow, I doubt you did them by hand:) Jul 13, 2015 at 20:24
• @martycohen: of course, to prevent errors, and allow me some sleep, I used Mathematica for expression manipulation. However, there is nothing that comes from a black box. For example, the argument to $\arcsin$ is $$2n^{-3/2} \left(1+\frac1n\right)^{-1/2} \left(1+\left(1-\frac1n\right)^{1/2}\right)^{-1}$$ which can be approached with the binomial theorem and series multiplication and composition (which are arduous, but not conceptually difficult).
– robjohn
Jul 14, 2015 at 0:33
• @martycohen: Then of course, composition with the series for $\arcsin$ which can be obtained by integrating the series for $(1-x^2)^{-1/2}$ which can be obtained using the binomial theorem. Again, arduous but not conceptually difficult.
– robjohn
Jul 14, 2015 at 0:33
• @martycohen: The differential operator for the Euler-Maclaurin Sum formula is the series for $\frac{x}{1-e^{-x}}$, which can be computed as the reciprocal of a power series. Just integrate the power series gotten from $\arcsin$ and apply the differential operator and we get the final series used to estimate the sum.
– robjohn
Jul 14, 2015 at 0:34
• @martycohen: the actual instruction I gave to Mathematica to compute the arcsine is Series[ArcSin[2/Sqrt[n (n + 1)]/(Sqrt[n] + Sqrt[n - 1])], {n, Infinity, 23}] I have also written a small package to handle the Euler-Maclaurin Sum Formula. It is based on what I said above.
– robjohn
Jul 14, 2015 at 0:36

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 $\approx1.57713$.