Question

If the value of $\frac{1}{2}+\frac{1}{3}-\frac{1}{3}(\frac{1}{2^3}+\frac{1}{3^3})+\frac{1}{5}(\frac{1}{2^5}+\frac{1}{3^5})-....... \infty$ is $\frac{a}{8}$ (where $a \in R$), then the value of $[a]$ is (where denotes greatest integer function)

Answer 1

6

Answer 2

The given series is $S = \frac{1}{2}+\frac{1}{3}-\frac{1}{3}(\frac{1}{2^3}+\frac{1}{3^3})+\frac{1}{5}(\frac{1}{2^5}+\frac{1}{3^5})-.......$

This can be written in summation notation as: $S = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{2n-1} \left(\frac{1}{2^{2n-1}} + \frac{1}{3^{2n-1}}\right)$

This can be split into two separate series: $S = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{2n-1} \left(\frac{1}{2}\right)^{2n-1} + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{2n-1} \left(\frac{1}{3}\right)^{2n-1}$

We recognize the Taylor series expansion for $\arctan(x)$: $\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{2n-1}}{2n-1}$, for $|x| \le 1$.

Using this, the first part of the series is $\arctan\left(\frac{1}{2}\right)$ and the second part is $\arctan\left(\frac{1}{3}\right)$. So, $S = \arctan\left(\frac{1}{2}\right) + \arctan\left(\frac{1}{3}\right)$.

We use the arctan addition formula: $\arctan(x) + \arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$, valid for $xy < 1$. Here, $x = \frac{1}{2}$ and $y = \frac{1}{3}$. Since $xy = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} < 1$, the formula applies.

$S = \arctan\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)}\right)$ $S = \arctan\left(\frac{\frac{3+2}{6}}{1-\frac{1}{6}}\right)$ $S = \arctan\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right)$ $S = \arctan(1)$

The principal value of $\arctan(1)$ is $\frac{\pi}{4}$. So, $S = \frac{\pi}{4}$.

We are given that $S = \frac{a}{8}$. Therefore, $\frac{\pi}{4} = \frac{a}{8}$. Solving for $a$: $a = 8 \times \frac{\pi}{4} = 2\pi$.

We need to find the value of $[a]$, which is the greatest integer function of $a$. $a = 2\pi$. Using the approximate value of $\pi \approx 3.14159$: $a \approx 2 \times 3.14159 = 6.28318$.

The greatest integer less than or equal to $6.28318$ is $6$. $[a] = [2\pi] = 6$.