The radius of convergence of the following power series is $ \sum_{n=0}^{\infty} \frac{(x-3)^n}{3^n n!} $
Power Series Radius of Convergence Calculation
To determine the radius of convergence (R) for the power series $ \sum_{n=0}^{\infty} \frac{(x-3)^n}{3^n n!} $, we apply the Ratio Test.
Ratio Test Application
The Ratio Test is used for power series of the form $ \sum a_n (x-c)^n $. We examine the limit $ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $, where $ a_n $ are the coefficients of the series terms excluding $ (x-c)^n $. The radius of convergence R is given by $ R = \frac{1}{L} $.
For the series $ \sum_{n=0}^{\infty} \frac{(x-3)^n}{3^n n!} $, the coefficients are $ c_n = \frac{1}{3^n n!} $. Thus, $ c_{n+1} = \frac{1}{3^{n+1} (n+1)!} $.
Calculate the limit L:
$ L = \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{1}{3^{n+1} (n+1)!}}{\frac{1}{3^n n!}} \right| $
Simplify the ratio:
$ L = \lim_{n \to \infty} \left| \frac{3^n n!}{3^{n+1} (n+1)!} \right| = \lim_{n \to \infty} \left| \frac{3^n \cdot n!}{3 \cdot 3^n \cdot (n+1) \cdot n!} \right| $
$ L = \lim_{n \to \infty} \left| \frac{1}{3(n+1)} \right| $
Evaluate the limit:
$ L = \frac{1}{3} \lim_{n \to \infty} \frac{1}{n+1} = \frac{1}{3} \cdot 0 = 0 $
Radius of Convergence Result
Since $ L = 0 $, the radius of convergence is:
$ \frac{1}{R} = 0 $
$ R = \infty $
Final Answer
The radius of convergence for the given power series is $ \infty $.
Important Questions from Infinite Series
- Consider the two series, $S_A$ and $S_B$, where
$$S_A = \sum_{n=1}^\infty \frac{n^2}{2^n}$$
$$S_B = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64} + \frac{1}{128} + \frac{1}{512} + \cdots$$
Which of the following statements is correct for the two given series? - The value of $\sum_{i=0}^{\infty} \sum_{j=1}^{\infty} 2^{-i} 3^{-j}$ is ______________ . (Answer in integer)
- Match each entry of List-1 with a suitable entry in List-2 and choose the correct option.
List-1 List-2 P The sum of the series $\sum_{n=1}^\infty \frac{1}{(n+2)(n+1)}$ is equal to I $\frac{3}{2}$ Q $\lim_{x \to 0} \left( \frac{3}{x^2} \int_0^x \sin(t) dt \right)$ is equal to II $1$ R Let $\frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx)$ be the Fourier series expansion of the function $f(x) = \frac{1}{2} \sin x - \frac{1}{2} \cos x + \frac{1}{\sqrt{2}} \sin 2x, x \in [0, 2\pi]$. Then, $\sum_{n=0}^\infty (a_n^2 + b_n^2)$ is equal to III $\frac{1}{2}$ The sum of the following infinite series is
$2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{8} + \frac{1}{9} + \frac{1}{16} + \frac{1}{27} + \dots$Consider the following two series
P: $\sum_{n=1}^{\infty} \frac{1}{n}$
Q: $\sum_{n=1}^{\infty} \frac{1}{n^2}$
Choose the correct option from the following
