Question

If \(R\) denotes the radius of convergence of the power series

• A. \(9R=4\)
• B. \(4R=9\)
• C. \(R=1\)
• D. \(2R=1\)

Hint:

For power series, the ratio test is often the fastest method to find the radius of convergence: \[ R=\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right| \] when the limit exists.

Answer 1

The Correct Option is B

Step 1: Identify the coefficient of the power series.
The given power series is
\[ \sum_{n=1}^{\infty} a_n x^n \] where
\[ a_n= \left( \frac{2\cdot4\cdot6\cdots2n} {2\cdot5\cdot8\cdots(3n-1)} \right)^2 \]
We need to find the radius of convergence \(R\).

Step 2: Use the ratio test formula for radius of convergence.
For a power series
\[ \sum a_n x^n, \] the radius of convergence is given by
\[ R= \lim_{n\to\infty} \left| \frac{a_n}{a_{n+1}} \right| \] provided the limit exists.

Step 3: Find the ratio \(\frac{a_n}{a_{n+1}}\).
We have
\[ a_n= \left( \frac{2\cdot4\cdot6\cdots2n} {2\cdot5\cdot8\cdots(3n-1)} \right)^2 \] and
\[ a_{n+1}= \left( \frac{2\cdot4\cdot6\cdots2n(2n+2)} {2\cdot5\cdot8\cdots(3n-1)(3n+2)} \right)^2 \]
Therefore,
\[ \frac{a_n}{a_{n+1}} = \left( \frac{3n+2}{2n+2} \right)^2 \]

Step 4: Take the limit as \(n\to\infty\).
\[ R= \lim_{n\to\infty} \left( \frac{3n+2}{2n+2} \right)^2 \]
Divide numerator and denominator by \(n\):
\[ R= \left( \frac{3+\frac{2}{n}} {2+\frac{2}{n}} \right)^2 \]
As
\[ n\to\infty, \] we get
\[ R= \left( \frac{3}{2} \right)^2 \] \[ R=\frac{9}{4} \]

Step 5: Match with the given options.
We obtained
\[ R=\frac{9}{4} \]
Multiplying both sides by \(4\),
\[ 4R=9 \]
Thus, option (B) is correct.

Step 6: Final conclusion.
Hence, the correct statement is
\[ \boxed{4R=9} \]