Question

Find the value of $1-\frac{1}{5}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{6n-5}-\frac{1}{6n-1}+....$

Answer 1

$\frac{\pi}{2\sqrt{3}}$

Answer 2

Let the given series be $S$. $S = 1-\frac{1}{5}+\frac{1}{7}-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+....$

The general term of the series can be written as $\left( \frac{1}{6n-5} - \frac{1}{6n-1} \right)$ for $n=1, 2, 3, \dots$. So, $S = \sum_{n=1}^{\infty} \left( \frac{1}{6n-5} - \frac{1}{6n-1} \right)$.

We can express this series as the value of a definite integral. Consider the function $f(x) = \sum_{n=1}^{\infty} \left( \frac{x^{6n-5}}{6n-5} - \frac{x^{6n-1}}{6n-1} \right)$. Then $S = f(1)$.

To find $f(x)$, we first find $f'(x)$: $f'(x) = \sum_{n=1}^{\infty} \left( x^{6n-6} - x^{6n-2} \right)$ $f'(x) = (1 - x^4) + (x^6 - x^{10}) + (x^{12} - x^{16}) + \dots$ This is a geometric series. $f'(x) = \sum_{n=0}^{\infty} x^{6n} - \sum_{n=0}^{\infty} x^{6n+4}$ $f'(x) = \frac{1}{1-x^6} - \frac{x^4}{1-x^6} = \frac{1-x^4}{1-x^6}$.

We can simplify this expression by factoring the numerator and denominator: $1-x^4 = (1-x^2)(1+x^2)$ $1-x^6 = (1-x^2)(1+x^2+x^4)$ So, $f'(x) = \frac{(1-x^2)(1+x^2)}{(1-x^2)(1+x^2+x^4)} = \frac{1+x^2}{1+x^2+x^4}$.

Now, we need to find $f(1) = \int_0^1 f'(x) dx$. $S = \int_0^1 \frac{1+x^2}{1+x^2+x^4} dx$.

To evaluate this integral, we can divide the numerator and denominator by $x^2$: $S = \int_0^1 \frac{\frac{1}{x^2}+1}{\frac{1}{x^2}+1+x^2} dx$ We can rewrite the denominator as $x^2+\frac{1}{x^2}+1 = \left(x-\frac{1}{x}\right)^2 + 2 + 1 = \left(x-\frac{1}{x}\right)^2 + 3$. So, $S = \int_0^1 \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+3} dx$.

Let $u = x-\frac{1}{x}$. Then $du = \left(1+\frac{1}{x^2}\right)dx$. When $x=1$, $u = 1-\frac{1}{1} = 0$. When $x \to 0^+$, $u \to 0 - \infty = -\infty$.

So the integral becomes: $S = \int_{-\infty}^{0} \frac{du}{u^2+3}$ This is a standard integral of the form $\int \frac{1}{a^2+x^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right)$. Here $a^2=3$, so $a=\sqrt{3}$. $S = \left[ \frac{1}{\sqrt{3}} \arctan\left(\frac{u}{\sqrt{3}}\right) \right]_{-\infty}^{0}$ $S = \frac{1}{\sqrt{3}} \left( \arctan\left(\frac{0}{\sqrt{3}}\right) - \lim_{u \to -\infty} \arctan\left(\frac{u}{\sqrt{3}}\right) \right)$ $S = \frac{1}{\sqrt{3}} \left( 0 - \left(-\frac{\pi}{2}\right) \right)$ $S = \frac{1}{\sqrt{3}} \left( \frac{\pi}{2} \right) = \frac{\pi}{2\sqrt{3}}$.

The final answer is $\frac{\pi}{2\sqrt{3}}$.