Question

Consider the infinite series \[ (P):\ \sum_{n=2}^{\infty}\frac{1}{(n\log n)^{1/n}}, \qquad (Q):\ \sum_{n=1}^{\infty}\frac{n^n}{(2n)!}. \] Then which one of the following statements is correct?

• A. Series $(P)$ and $(Q)$ both converge
• B. Series $(P)$ converges and series $(Q)$ diverges
• C. Series $(P)$ and $(Q)$ both diverge
• D. Series $(P)$ diverges and series $(Q)$ converges

Hint:

Before applying heavy tests, check the $n$th-term limit. If $a_n\nrightarrow 0$, the series diverges immediately. For factorial versus power terms, the ratio test is typically decisive.

Answer 1

The Correct Option is D

Step 1 (Series $P$): Test the $n$th term.
\[ a_n=\frac{1}{(n\log n)^{1/n}}=\exp\!\left(-\frac{\ln n+\ln\ln n}{n}\right). \] Since $\displaystyle \frac{\ln n}{n}\to 0$ and $\displaystyle \frac{\ln\ln n}{n}\to 0$, we have \[ (n\log n)^{1/n}\to e^{0}=1 \Rightarrow a_n\to 1\neq 0. \] Hence the necessary condition for convergence fails; therefore \(\sum a_n\) diverges.

Step 2 (Series $Q$): Ratio test.
Let \(b_n=\dfrac{n^n}{(2n)!}\). Then \[ \frac{b_{n+1}}{b_n} =\frac{(n+1)^{n+1}}{(2n+2)!}\cdot\frac{(2n)!}{n^n} = \frac{(n+1)^{n+1}}{n^n(2n+2)(2n+1)} = \frac{(1+\tfrac1n)^n}{2(2n+1)}. \] As \(n\to\infty\), \((1+\tfrac1n)^n\to e\) and \(2(2n+1)\sim 4n\), so \[ \lim_{n\to\infty}\frac{b_{n+1}}{b_n}=\lim_{n\to\infty}\frac{e}{4n}=0<1. \] By the ratio test, \(\sum b_n\) converges. Final Answer: \(\boxed{\text{Option (D)}}\)