MHB Summation of an infinite series

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The discussion focuses on proving the equality of the infinite series $\sum_{n=0}^{\infty} (-1)^{n} \arctan \left( \frac{1}{2n+1} \right)$ and $\arctan \left( \text{tanh} \left( \frac{\pi}{4} \right) \right)$. Participants suggest using the complex logarithm representation of the inverse tangent and trigonometric identities to transform the series. A specific formula involving the sum of arctangents is introduced, leading to the conclusion that the series can be expressed in terms of $\tan^{-1}\left(\frac{1}{2(2n+1)^{2}}\right)$. Ultimately, applying the derived formula confirms the original equality.
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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.
 
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Random Variable said:
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
 
CaptainBlack said:
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.
 
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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