Question

If the sum of the series \[1+\dfrac{2}{x}+\dfrac{4}{{{x}^{2}}}+\dfrac{8}{{{x}^{3}}}+...\infty \] is a finite number, then
A. \[x>2\]
B. \[x>-2\]
C. $x>\dfrac{1}{2}$
D. None of these

Answer 1

Hint: In this question, we are to find the sum of infinite terms of the given series. By using that we can able to find the limit/condition for the sum to be a finite number.

b>Formula Used:The sum of the infinite terms in the G.P series is calculated by
 ${{S}_{\infty }}=\dfrac{a}{1-r}$ where $r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{\infty }}$ is the sum of the infinite terms of the series; $a$ is the first term in the series and $r$ is the common ratio.

Complete step by step solution: The given series is
\[1+\dfrac{2}{x}+\dfrac{4}{{{x}^{2}}}+\dfrac{8}{{{x}^{3}}}+...\infty \]
Here the first term \[a=1\];
And the common ratio \[r=\dfrac{\dfrac{4}{{{x}^{2}}}}{\dfrac{2}{x}}=\dfrac{2}{x}\]
So, the sum of infinite terms in the series is
\[\begin{align}
  & {{S}_{\infty }}=\dfrac{a}{1-r} \\
 & \text{ }=\dfrac{1}{1-\dfrac{2}{x}} \\
 & \text{ }=\dfrac{x}{x-2} \\
\end{align}\]
Since the sum has $x-2$ in the denominator, it must be greater than zero to become a finite number. I.e.,
\[\begin{align}
  & x-2>0 \\
 & \Rightarrow x>2 \\
\end{align}\]

Option ‘A’ is correct

Note: Here the given series is geometric series. So, by using the appropriate formula, the sum of infinite terms is calculated. In order to be a finite number, the denominator f the rational number must be greater than zero. We can also use the common ratio for finding the condition to be a finite number.