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summation of an infinite series

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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
Jan 26, 2012
890
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
Jan 31, 2012
253
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
Apr 2, 2012
9
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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