Discussion
The radius of convergence of the series \( \sum_{n=0}^{\infty} \frac{(2n)!}{(n!)^{2}} (z - 3i)^{n} \) is :
Answer: D
The radius of convergence is \( \frac{1}{4} \).
By the ratio test, with \( a_{n} = \frac{(2n)!}{(n!)^{2}} \):
\[ \begin{aligned} \frac{a_{n+1}}{a_{n}} &= \frac{(2n+2)(2n+1)}{(n+1)^{2}} \\ &\to 4 \text{ as } n \to \infty \end{aligned} \]
The series converges when \( 4|z - 3i| \lt 1 \), that is within a distance \( R = 1/4 \) of the centre \( 3i \). The centre does not affect the radius.
Infinity would need the ratio to tend to 0, as for \( \sum z^{n}/n! \). \( 1/2 \) would be the radius if the series were in powers of \( (z - 3i)^{2} \), and 2 is that value inverted; neither fits a series in \( (z - 3i)^{n} \).
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development