Analysis of Infinite Series Convergence

The question requires determining the convergence or divergence of two infinite series, (P) and (Q).

Series (P): \sum_{n=2}^{\infty} \frac{1}{(n \log n)^{1/n}}$

Let the terms of the series be $a_n = \frac{1}{(n \log n)^{1/n}}$. We apply the Test for Divergence, which states that if $\lim_{n \to \infty} a_n \neq 0$, the series diverges.

1. Consider the limit of the denominator: $L = \lim_{n \to \infty} (n \log n)^{1/n}$.
2. To evaluate $L$, we examine its natural logarithm:

   $\log L = \lim_{n \to \infty} \frac{\log(n \log n)}{n} = \lim_{n \to \infty} \frac{\log n + \log(\log n)}{n}$

3. Applying L'Hopital's Rule:

   $\log L = \lim_{n \to \infty} \frac{\frac{1}{n} + \frac{1}{n \log n}}{1} = \lim_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n \log n}\right) = 0 + 0 = 0$.

4. Since $\log L = 0$, we have $L = e^0 = 1$.
5. Therefore, the limit of the terms is:

   $\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{(n \log n)^{1/n}} = \frac{1}{L} = \frac{1}{1} = 1$.

6. Because $\lim_{n \to \infty} a_n = 1 \neq 0$, series (P) diverges.

Series (Q): \sum_{n=1}^{\infty} \frac{n^n}{(2n)!}$

Let the terms of the series be $b_n = \frac{n^n}{(2n)!}$. We apply the Ratio Test to determine convergence.

1. Calculate the ratio $\frac{b_{n+1}}{b_n}$:

   $\frac{b_{n+1}}{b_n} = \frac{(n+1)^{n+1}}{(2(n+1))!} \cdot \frac{(2n)!}{n^n} = \frac{(n+1)^{n+1}}{(2n+2)!} \cdot \frac{(2n)!}{n^n}$

2. Simplify the expression:

   $= \frac{(n+1)^{n+1}}{(2n+2)(2n+1)(2n)!} \cdot \frac{(2n)!}{n^n} = \frac{(n+1)^{n+1}}{(2n+2)(2n+1) n^n}$

   $= \frac{(n+1)^n (n+1)}{2(n+1)(2n+1) n^n} = \frac{(n+1)^n}{2(2n+1) n^n}$

   $= \frac{1}{2(2n+1)} \left(\frac{n+1}{n}\right)^n = \frac{1}{2(2n+1)} \left(1 + \frac{1}{n}\right)^n$

3. Evaluate the limit as $n \to \infty$:

   $\lim_{n \to \infty} \left| \frac{b_{n+1}}{b_n} \right| = \lim_{n \to \infty} \left( \frac{1}{2(2n+1)} \left(1 + \frac{1}{n}\right)^n \right)$

   $= \left( \lim_{n \to \infty} \frac{1}{2(2n+1)} \right) \cdot \left( \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \right)$

   $= 0 \cdot e = 0$.

4. Since the limit is $0 < 1$, series (Q) converges by the Ratio Test.

Conclusion

Series (P) diverges and series (Q) converges.