Series Convergence Analysis
The given series is $\sum_{m=0}^{\infty} \frac{1}{4^m}(x-1)^{2m}$. We need to find the interval for $x$ where this series converges.
Rewrite the Series
The series can be rewritten by combining terms:
$ \sum_{m=0}^{\infty} \frac{1}{4^m}(x-1)^{2m} = \sum_{m=0}^{\infty} \left(\frac{(x-1)^2}{4}\right)^m $Identify Geometric Series
This is a geometric series of the form $\sum_{m=0}^{\infty} r^m$, where the common ratio is:
$ r = \frac{(x-1)^2}{4} $Apply Convergence Condition
A geometric series converges if and only if the absolute value of its common ratio is less than 1:
$ |r| < 1 $ $ \left|\frac{(x-1)^2}{4}\right| < 1 $Solve for x
Since $(x-1)^2$ is always non-negative, the absolute value is redundant:
$ \frac{(x-1)^2}{4} < 1 $Multiply both sides by 4:
$ (x-1)^2 < 4 $Take the square root of both sides:
$ \sqrt{(x-1)^2} < \sqrt{4} $ $ |x-1| < 2 $This inequality means:
$ -2 < x-1 < 2 $Add 1 to all parts of the inequality:
$ -2 + 1 < x-1 + 1 < 2 + 1 $ $ -1 < x < 3 $Conclusion
The series converges for the interval $-1 < x < 3$. This corresponds to Option B.
Important Questions from Infinite Series
- Consider the following series:
(i) $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$
(ii) $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$
(iii) $\sum_{n=1}^{\infty} \frac{1}{n!}$ - The sum of the following infinite series is:
$ \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + ... $ - The series
$\sum_{n=0}^{r} q^n = 1 + q + q^2 + \dots$ has the sum: The value of the series $1+ \sin x + \cos^2 x + \sin^3 x + \dots$ at $x = \frac{ \pi}{4}$ is __________.
The sum of the infinite geometric series $1+\frac{1}{3}+\frac{1}{3^2} + \frac{1}{3^3} + ...$ (rounded off to one decimal place) is____.
