# The closed form of $\sum_{n=0}^{\infty} \arcsin\bigl(\frac{1}{e^n}\bigr)$

In my study on some type of integrals I met the series below that I don't how to approach it.
Of course, one of the obvious questions is: does it have a closed form? Before answering that,
I need to learn how to tackle them, the proper tools to employ. Any help on this series is very welcome. The use of $\arcsin(x)$ series expansion wasn't fruitful.

$$\sum_{n=0}^{\infty} \arcsin\left(\frac{1}{e^n}\right)$$

that more generally can be considered as

$$\sum_{n=0}^{\infty} \arcsin\left(x^n\right)$$

• Numerically, $\sum_{n=0}^{\infty} \arcsin\left(\frac{1}{e^n}\right)\approx 2.1620693376$. A quick inverse symbolic calculator search turned up nothing... – David H May 1 '14 at 16:50
• There isn't any. – Lucian May 1 '14 at 17:13
• See here for a similar question. – Lucian May 1 '14 at 17:26
• @Lucian Thanks. I discussed that question in chat, but this one seems much harder. – user 1357113 May 1 '14 at 17:35
• Personally, I think it's not merely “harder”, but simply non-existent. By differentiating the arctangent sum, se get a familiar-looking quadratic series, but in this case we have a problematic radical sign in the denominator. – Lucian May 1 '14 at 17:50

While I doubt a closed form exists, using Lucian's idea, we can obtain a series with much better convergence.

First, we separate the terms with $n=0$, which is simply $\frac{\pi}{2}$ and consider the series:

$$f(x)=\sum_{n=1}^\infty \arcsin x^n, \qquad |x|<1$$

Differentiating by term, we obtain:

$$f'(x)=\sum_{n=1}^\infty \frac{n x^{n-1}}{\sqrt{1-x^{2n}}}$$

The series converges absolutely for $|x|<1$. Now we expand the denominator:

$$\frac{1}{\sqrt{1-x^{2n}}}=\sum_{k=0}^\infty \frac{\Gamma \left(k+\frac{1}{2} \right)}{\sqrt{\pi}~k!} x^{2nk}$$

We get a double series for the derivative:

$$f'(x)=\sum_{k=0}^\infty \frac{\Gamma \left(k+\frac{1}{2} \right)}{\sqrt{\pi}~k!} \sum_{n=1}^\infty n x^{2kn+n-1}$$

Integrating by term and using the condition $f(0)=0$, we have:

$$f(x)=\sum_{k=0}^\infty \frac{\Gamma \left(k+\frac{1}{2} \right)}{\sqrt{\pi}~k!} \sum_{n=1}^\infty \frac{n x^{(2k+1)n}}{(2k+1)n}=\sum_{k=0}^\infty \frac{\Gamma \left(k+\frac{1}{2} \right)}{\sqrt{\pi}~k!~(2k+1)} \frac{x^{2k+1}}{1-x^{2k+1}}$$

The advantage of this new series representation is that it converges faster and also has rational terms (for rational $x$). It's clear if we use another representation for the general term:

$$\frac{\Gamma \left(k+\frac{1}{2} \right)}{\sqrt{\pi}~k!}=\frac{(2k)!}{4^k~k!^2}$$

So finally we have:

$$\sum_{n=0}^{\infty} \arcsin x^n=\frac{\pi}{2}+\sum_{k=0}^\infty \frac{(2k)!}{4^k~k!^2 ~(2k+1)} \frac{x^{2k+1}}{1-x^{2k+1}}$$

For the case presented by the OP we have:

$$\sum_{n=0}^{\infty} \arcsin\left(\frac{1}{e^n}\right)=\frac{\pi}{2}+f\left( \frac{1}{e} \right)$$

$$\sum_{n=0}^{\infty} \arcsin\left(\frac{1}{e^n}\right)=\frac{\pi}{2}+\sum_{k=0}^\infty \frac{(2k)!}{4^k~k!^2 ~(2k+1)} \frac{1}{e^{2k+1}-1}$$

Using just $5$ terms of the series, we obtain the value $2.16205929\dots$, which after rounding off gives $5$ correct digits after the decimal point $2.16206\dots$. While $5$ terms of the original series gives much less accurate value $2.158\dots$.