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: Write the coefficient.
The term in $x^n$ has coefficient $a_n=\dfrac{(n!)^4}{(2n)!\,\log n}$.

Step 2: Plan with the ratio test.
The radius $R$ comes from $\dfrac1R=\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$.

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

Step 4: Find the size.
For large $n$, $\frac{(n+1)^4}{(2n+2)(2n+1)}\sim \frac{n^4}{4n^2}=\frac{n^2}{4}$, while $\frac{\log n}{\log(n+1)}\to 1$.

Step 5: Take the limit.
So the ratio grows like $\frac{n^2}{4}\to\infty$. That gives $\frac1R=\infty$.

Step 6: Read off $R$.
Hence the radius of convergence is $0$.
\[ \boxed{0.0} \]