Summation of an infinite series

[polygamma]

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]

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

 

[polygamma]

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.

 

[fredoniahead]

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

 

[CellCoree]

I would first clarify the concept of an infinite series for those who may not be familiar with it. An infinite series is a sum of an infinite number of terms, each of which follows a specific pattern. In this case, the series starts at n=0 and continues indefinitely, with each term being the arctan of 1 divided by an odd number (2n+1) raised to the power of (-1)^n.

To show that this infinite series is equal to the arctan of the hyperbolic tangent of pi/4, we can use the identity for the arctan of a hyperbolic function:

arctan(tanh(x)) = x

Therefore, we can rewrite the right side of the equation as:

arctan(tanh(pi/4)) = pi/4

Now, we need to show that the left side of the equation, the infinite series, also equals pi/4. To do this, we can use the formula for the arctan of a fraction:

arctan(x) = 1/2 * ln((1+x)/(1-x))

Substituting in the terms in our infinite series, we get:

arctan(1/(2n+1)) = 1/2 * ln((2n+2)/(2n))

Using the properties of logarithms, we can simplify this to:

arctan(1/(2n+1)) = ln(sqrt((2n+2)/(2n)))

Now, we can rewrite our infinite series as:

ln(sqrt((2*1)/(2*0))) - ln(sqrt((2*3)/(2*2))) + ln(sqrt((2*5)/(2*4))) - ln(sqrt((2*7)/(2*6))) + ...

= ln(sqrt((2n+2)/(2n)))

= ln(sqrt((2(n+1))/(2n)))

= ln(sqrt(1+1/n))

As n approaches infinity, the value of this expression approaches ln(sqrt(1)) = 0.

Therefore, the left side of our equation, the infinite series, is equal to 0.

This confirms that:

arctan(tanh(pi/4)) = pi/4

and thus,

$\displaystyle \sum_{n=0}^{\infty} (-1)^{n} \