Find this sum, if you can

Rewriting as $\displaystyle 1 + \sum_{n=1}^{\infty}- \frac{1}{(4n-1)3^{2n-1}}+\frac{1}{(4n+1)3^{2n}} =$

$\displaystyle 1+\sum_{n=1}^{\infty}\int_{0}^{1}\frac{y^{4n-2}}{9^n}dy+\frac{y^{4n}}{9^n}dy =$

Then since the integral and sum are with respect different variables, they can be switched and the sums evaluated with $0<y<1$

$\displaystyle 1+\int_{0}^{1}( \sum_{n=1}^{\infty}\frac{-y^{4n-2}}{9^n}+\frac{y^{4n}}{9^n})dy =$

$\displaystyle 1+\int_{0}^{1}\frac{y^4-3y^2}{9-y^4}dy = \frac{\pi}{2\sqrt{3}}$

$\displaystyle\frac{\sqrt{12}\pi}{2\sqrt{3}} = \bf\boxed{\pi}$

Consider the Macluarin series for $\tan^{-1} x$ .

$\begin{aligned} \tan^{-1} x & = x - \frac {x^3}3 + \frac {x^5}5 - \frac {x^7}7 + ... & \small {\color{#3D99F6}\text{Putting }x=\frac 1{\sqrt 3}} \\ \tan^{-1} \frac 1{\sqrt 3} & = \frac 1{\sqrt 3} - \frac 1{3(\sqrt 3)^3} + \frac 1{5(\sqrt 3)^5} - \frac 1{7(\sqrt 3)^7} + ... & \small {\color{#3D99F6}\text{Multiplying both sides with } \sqrt 3} \\ \sqrt 3 \tan^{-1} \frac 1{\sqrt 3} & = 1 - \frac 1{3\cdot 3} + \frac 1{5\cdot 3^2} - \frac 1{7\cdot 3^3} + ... \\ & = S \end{aligned}$

$\implies \sqrt{12} S = \sqrt{12} \cdot \sqrt 3 \tan^{-1} \dfrac 1{\sqrt 3} = 6 \cdot \dfrac \pi 6 = \pi \approx \boxed{3.14}$

I did it using complex analysis of taylor seires for ln(1-x) . Please download and zoom in the picture to see clearly and the picture quality is HDR so it is quite clear.

{: .cente

We can get this sum through some manipulation of the Maclaurin series for $\arctan x$ (Derivation below), which is $\arctan x = x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots.$ Dividing both sides by $x$ , we get $\frac{\arctan x}{x}=1-\frac{x^2}{3}+\frac{x^4}{5}-\cdots.$ Finally, by substituting $x=\frac{1}{\sqrt{3}}$ , we get $\begin{aligned} S&=&1-\frac{1}{3\times3}+\frac{1}{5\times3^2}-\frac{1}{7\times3^3}+\cdots\\&=&\frac{\arctan\left(\frac{1}{\sqrt{3}}\right)}{\frac{1}{\sqrt{3}}}\\ &=&\frac{\sqrt{3}\pi}{6} \end{aligned}$ Hence, $\sqrt{12}\times S=\pi.$ QED.

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Do note that this solution works because the substitution we used here is in the range $(-1,1)$ (this is also known as the radius of convergence of the series). Now the derivation of the Maclaurin series for $\arctan x$ . Consider $\begin{aligned} \arctan x&=&\int_0^x\frac{1}{1+t^2} dt\\ &=&\int_0^x \left(1-x^2+x^4-x^6+\cdots\right) dt\ \ \ \ (\textrm{By Binomial expansion, and } |t|<1)\\ &=&x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots, |x|<1. \end{aligned}$ The derivation is complete.