An arctan series

Calculus (Integrals, Series)
Post Reply
User avatar
Tolaso J Kos
Administrator
Administrator
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

An arctan series

#1

Post by Tolaso J Kos »

Prove that:

$$\sum\limits_{n = 1}^\infty {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}} = \ln 3 - \frac{\pi }{4}$$
Imagination is much more important than knowledge.
r9m
Posts: 59
Joined: Thu Dec 10, 2015 1:58 pm
Location: India
Contact:

Re: An arctan series

#2

Post by r9m »

\begin{align*}S&=\sum\limits_{n=1}^{\infty} \arctan \frac{10n}{(3n^2+2)(9n^2-1)} \\&= \sum\limits_{n=1}^{\infty} \arg \left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(\frac{(3n^2+2)(9n^2-1)+10in}{27n^4\left(1+\frac{2}{3n^2}\right)\left(1-\frac{1}{9n^2}\right)}\right)\end{align*}

The products in the denominator $\displaystyle \prod\limits_{n=1}^{\infty}\left(1+\frac{2}{3n^2}\right)$ and $\displaystyle \prod\limits_{n=1}^{\infty}\left(1-\frac{1}{9n^2}\right)$ can be ignored as they are convergent and real.

So, \begin{align*}S&= \arg \prod\limits_{n=1}^{\infty}\left(\frac{(3n^2+2)(9n^2-1)+10in}{27n^4}\right)\end{align*}

Using the factorization of the numerator,

$$(3n^2+2)(9n^2-1)+10in = (n-i)(3n+i)(3n+i+1)(3n+i-1)$$ we get,

$$S = \arg \prod\limits_{n=1}^{\infty}\frac{\left(1+\frac{i}{3n}\right)\left(1+\frac{i+1}{3n}\right)\left(1+\frac{i-1}{3n}\right)}{\left(1+\frac{i}{n}\right)}$$

Using, $\displaystyle \frac{1}{\Gamma(z)} = ze^{\gamma z}\prod\limits_{n=1}^{\infty}\left(1+\frac{z}{n}\right)e^{-z/n}$ at $\displaystyle z = i,\frac{i}{3},\frac{i+1}{3},\frac{i-1}{3}$ we may rewrite it as,

$$S = \arg \frac{-\Gamma(i)}{\Gamma\left(\frac{i}{3}\right)\Gamma\left(\frac{i+1}{3}\right)\Gamma\left(\frac{i-1}{3}\right)}$$

On the other hand we have Gauss-Legendre Triplication Formula

$\displaystyle \Gamma(3z) = \frac{1}{2\pi}3^{3z - \frac{1}{2}}\Gamma\left(z\right)\Gamma\left(z+\frac{1}{3}\right)\Gamma\left(z+\frac{2}{3}\right)$,

which at $z = \dfrac{i-1}{3}$ gives:

$$\Gamma\left(\frac{i-1}{3}\right)\Gamma\left(\frac{i}{3}\right)\Gamma\left(\frac{i+1}{3}\right) = 2\pi 3^{-i+\frac{3}{2}}\Gamma(i-1)$$

Hence, $$S = \arg \frac{-3^{i}\Gamma(i)}{\Gamma(i-1)} = \arg (3^{i}(1-i)) = \log 3 - \frac{\pi}{4}$$

Here's a link to another solution.
User avatar
Tolaso J Kos
Administrator
Administrator
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

Re: An arctan series

#3

Post by Tolaso J Kos »

Well , well, well I knew that this series was quite popular for I have seen it in many fora including mathematica.gr. So, I thought I'd add it in our collection.

Thank you r9m , both for your solution and the link.
Imagination is much more important than knowledge.
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 14 guests