Something Beyond Reach

Since $4(r + 1) + 1 = 4r + 5$ , the denominators in this series will just be the sequence of odd natural numbers starting at $1$ but with alternating signs, i.e., $1, -3, 5, -7, 9, .....$

Now the series for the inverse tangent function is

$\tan^{-1}(x) = x - \dfrac{x^{3}}{3} + \dfrac{x^{5}}{5} - \dfrac{x^{7}}{7} + .....$

But as $\tan^{-1}(1) = \dfrac{\pi}{4}$ we see that

$\dfrac{\pi}{4} = 1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + ......$ ,

and so the desired sum is $\dfrac{\pi}{4} = \boxed{0.785}$ to 3 decimal places.

Let required Limit as ' L ' . Now This expression is look like telescopic sum , but it is not. we realise after writing 2-3 terms. So what is it ? Yes actually , $1/4r+1$ mean's integration of ${ x }^{ 4r }$ in (0,1) . So required sum is analogous to : $\displaystyle{L=\lim _{ n\rightarrow \infty }{ \sum _{ r=0 }^{ n }{ \int _{ 0 }^{ 1 }{ ({ x }^{ 4r }-{ x }^{ 4r+2 })dx } } } \\ L=\int _{ 0 }^{ 1 }{ (\lim _{ n\rightarrow \infty }{ \sum _{ 0 }^{ n }{ { x }^{ 4r } } } )(1-{ x }^{ 2 })dx } }$ Now Using GP's sum formula , we get $\displaystyle{L=\int _{ 0 }^{ 1 }{ (\cfrac { 1 }{ 1-{ x }^{ 4 } } )(1-{ x }^{ 2 })dx } \\ L=\int _{ 0 }^{ 1 }{ \cfrac { dx }{ 1+{ x }^{ 2 } } } ={ \left[ \tan ^{ -1 }{ x } \right] }_{ 0 }^{ 1 }\\ L=\cfrac { \pi }{ 4 } }$

Let $f$ be a function defined on $[0,1]$ by :

$\displaystyle f(x) = \sum_{r=0}^{\infty} (\frac{x^{4r+1}}{4r+1} - \frac{x^{4r+3}}{4r+3} )$

Where we seek $f(1)$ and we have $f(0) = 0$ .Differentiate:

$\displaystyle f'(x) = \sum_{r=0}^{\infty} (x^{4r} - x^{4r+2} )$

$\displaystyle = \sum_{r=0}^{\infty} (x^{4r} - x^2x^{4r})$

$\displaystyle = (1-x^2) \sum_{r=0}^{\infty} (x^4)^r$

$\displaystyle = \frac{1-x^2}{1-x^4} = \frac{1}{x^2+1}$

We have : $f(1) = f(1)-f(0)$

Fundamental Theorem of Calculus :

$\displaystyle => f(1) = \int_0^1 f'(x) dx = \int_0^1 \frac{dx}{x^2+1}$

$\displaystyle \boxed{=\frac{\pi}{4} }$

for those who r interested in higher mathematics,use the digamma function...jus go to page 9 of this link which is pdf we have a direct formmula for such sums(but i like the arctanx series approach)

https://www.math.washington.edu/~morrow/336_10/papers/joel.pdf

We get $1-\frac{1}{3} + \frac{1}{5} - \frac{1}{7} +....$

Now $tan^{-1} x =$ integration of $\frac{1}{1+x^2} dx$ = integration of $(1+x^2)^{-1} dx$ = integration of $1+ (x^2)^1 \frac{(-1)}{1!} + (x^2)^2 \frac{(-1)(-2)}{2!} + (x^2)^3 \frac{(-1)(-2)(-3)}{3!} + .... dx$ = integration of $1- x^2 + x^4 - x^6 +... dx= x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ...$

Sub in $x= 1$ we have $tan^{-1} 1 = 1 -\frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ...$ which is the given series.

So the above series is equal to $tan^{-1} 1 = \frac{\pi}{4} = \boxed{0.785}$ .

using fox theorem