# summation of an infinite series

#### Random Variable

##### Well-known member
MHB Math Helper
Show that $\displaystyle \sum_{n=0}^{\infty} (-1)^{n} \arctan \left( \frac{1}{2n+1} \right) = \arctan \Bigg( \text{tanh} \Big( \frac{\pi}{4} \Big) \Bigg)$.

I'm tempted to give a hint (or two) right off the bat. But I'll wait.

#### CaptainBlack

##### Well-known member
Show that $\displaystyle \sum_{n=0}^{\infty} (-1)^{n} \arctan \left( \frac{1}{2n+1} \right) = \arctan \Bigg( \text{tanh} \Big( \frac{\pi}{4} \Big) \Bigg)$.

I'm tempted to give a hint (or two) right off the bat. But I'll wait.
Would the hınt dırect one ın the towards of the exponentıal representatıon of tan and the use of logs by any chance?

CB

#### Random Variable

##### Well-known member
MHB Math Helper
Would the hınt dırect one ın the towards of the exponentıal representation of tan and the use of logs by any chance?

CB

Yes, I was going to suggest representing the inverse tangent using the complex logarithm, but in a simpler way than it's usually represented. But I was also going to suggest using a trig identity to first write the series in a different form so that's it's not an alternating series.

#### quantaentangled

##### New member
Hi RV. Cool problem.

$$\frac{\pi}{4}-\tan^{-1}(1/3)+tan^{-1}(1/5)-\tan^{-1}(1/7)+\cdot\cdot\cdot$$

$$\left(\frac{\pi}{4}+\tan^{-1}(1/5)+\tan^{-1}(1/9)+\cdot\cdot\cdot\right) -\left(\tan^{-1}(1/3)+tan^{-1}(1/7)+\tan^{-1}(1/11)+\cdot\cdot\cdot \right)$$

This can now be written as:

$$\sum_{n=0}^{\infty}\left[\tan^{-1}\left(\frac{1}{4n+1}\right)-\tan^{-1}\left(\frac{1}{4n+3}\right)\right]$$
$$\sum_{n=0}^{\infty}\left[\tan^{-1}(4n+3)-\tan^{-1}(4n+1)\right]$$

$$=\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{1}{8n^{2}+8n+2}\right)$$

$$=\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{1}{2(2n+1)^{2}}\right)$$

Now, we can use the formula:

$$\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{2xy}{(2n+1)^{2}-x^{2}+y^{2}}\right)=\tan^{-1}\left(\tan(\frac{{\pi}x}{2})\tanh(\frac{{\pi}y}{2})\right)$$

Now, let $$x=y=1/2$$ in the formula and we arrive at:

$$\sum_{n=0}^{\infty}\tan^{-1}\left(\frac{1}{2(2n+1)^{2}}\right)=\tan^{-1}\left(\tanh(\frac{\pi}{4})\right)$$

The formula above is derived by using the argument of

$$\cos(\pi z)=\prod_{k=0}^{\infty}\left(1-\frac{4z^{2}}{(2k+1)^{2}}\right)$$

$$\cos(\pi z)=\cos(\pi x)\cosh(\pi y)+\sin(\pi x)\sinh(\pi y)i$$.

and by noting that $$\text{arg}\prod(a+bi)=\sum \tan^{-1}\left(\frac{b}{a}\right)$$

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