Question

The radius of convergence of the series \( \displaystyle \sum_{n=1}^{\infty} \frac{(n!)^4}{(2n)!}(\log_e n)^{-1}x^n \) is rounded off to one decimal place.

Hint:

If the ratio \( \left|\frac{a_{n+1}}{a_n}\right| \to \infty \), then the radius of convergence of \( \sum a_nx^n \) is \(0\).

Answer 1

Correct Answer: 0

Step 1: Identify the coefficient of \(x^n\).
\[ a_n=\frac{(n!)^4}{(2n)! \log_e n} \]

Step 2: Use ratio test for radius of convergence.
For the power series \( \sum a_n x^n \), radius of convergence is obtained from
\[ \frac{1}{R}=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| \]

Step 3: Write the ratio.
\[ \frac{a_{n+1}}{a_n} = \frac{((n+1)!)^4}{(2n+2)!\log(n+1)} \cdot \frac{(2n)!\log n}{(n!)^4} \]

Step 4: Simplify factorial terms.
\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^4}{(2n+2)(2n+1)} \cdot \frac{\log n}{\log(n+1)} \]

Step 5: Find the limiting behaviour.
\[ \frac{(n+1)^4}{(2n+2)(2n+1)} \sim \frac{n^4}{4n^2} = \frac{n^2}{4} \]
Also,
\[ \frac{\log n}{\log(n+1)} \to 1 \]

Step 6: Evaluate the limit.
\[ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \infty \]

Step 7: Find the radius of convergence.
\[ \frac{1}{R}=\infty \] \[ R=0 \] \[ \boxed{0.0} \]