For which positive integers $k$ is the following series convergent? $\sum n_1\infty \frac{{(n!)}^2}{(kn)!}$

• A. For all $k\ge 1$
• B. For all $k\ge 2$
• C. For all $k\ge 3$
• D. For no positive integer $k$

Answer: (C)

Verified step by step guidance

1. Step 1: Begin by analyzing the given series: \( \sum_{n=1}^{\infty} \frac{(n!)^2}{(kn)!} \). The convergence of the series depends on the growth rate of the numerator \((n!)^2\) compared to the denominator \((kn)!\).
2. Step 2: Use the ratio test to determine convergence. The ratio test states that a series \( \sum a_n \) converges if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \). Here, \( a_n = \frac{(n!)^2}{(kn)!} \).
3. Step 3: Compute \( \frac{a_{n+1}}{a_n} \): Substitute \( a_{n+1} = \frac{((n+1)!)^2}{((k(n+1))!)} \) and \( a_n = \frac{(n!)^2}{(kn)!} \) into the ratio \( \frac{a_{n+1}}{a_n} \). Simplify using factorial properties, such as \( (n+1)! = (n+1) \cdot n! \).
4. Step 4: Analyze the growth of \( (n!)^2 \) versus \( (kn)! \) as \( n \to \infty \). For small values of \( k \), the denominator \( (kn)! \) grows slower than \( (n!)^2 \), causing the series to diverge. For larger values of \( k \), \( (kn)! \) grows faster, leading to convergence.
5. Step 5: Conclude that the series converges for \( k \geq 3 \). For \( k = 1 \) or \( k = 2 \), the growth of \( (kn)! \) is insufficient to counterbalance \( (n!)^2 \), and the series diverges. For \( k \geq 3 \), the denominator grows sufficiently fast to ensure convergence.