(P): $\sum_{n=2}^{\infty} \frac{1}{(n \log n)^{1/n}}$ (Q): $\sum_{n=1}^{\infty} \frac{n^n}{(2n)!}$
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.
- Consider the limit of the denominator: $L = \lim_{n \to \infty} (n \log n)^{1/n}$.
- To evaluate $L$, we examine its natural logarithm:
$\log L = \lim_{n \to \infty} \frac{\log(n \log n)}{n} = \lim_{n \to \infty} \frac{\log n + \log(\log n)}{n}$
- Applying L'Hopital's Rule:
$\log L = \lim_{n \to \infty} \frac{\frac{1}{n} + \frac{1}{n \log n}}{1} = \lim_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n \log n}\right) = 0 + 0 = 0$.
- Since $\log L = 0$, we have $L = e^0 = 1$.
- Therefore, the limit of the terms is:
$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{(n \log n)^{1/n}} = \frac{1}{L} = \frac{1}{1} = 1$.
- 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.
- Calculate the ratio $\frac{b_{n+1}}{b_n}$:
$\frac{b_{n+1}}{b_n} = \frac{(n+1)^{n+1}}{(2(n+1))!} \cdot \frac{(2n)!}{n^n} = \frac{(n+1)^{n+1}}{(2n+2)!} \cdot \frac{(2n)!}{n^n}$
- Simplify the expression:
$= \frac{(n+1)^{n+1}}{(2n+2)(2n+1)(2n)!} \cdot \frac{(2n)!}{n^n} = \frac{(n+1)^{n+1}}{(2n+2)(2n+1) n^n}$
$= \frac{(n+1)^n (n+1)}{2(n+1)(2n+1) n^n} = \frac{(n+1)^n}{2(2n+1) n^n}$
$= \frac{1}{2(2n+1)} \left(\frac{n+1}{n}\right)^n = \frac{1}{2(2n+1)} \left(1 + \frac{1}{n}\right)^n$
- Evaluate the limit as $n \to \infty$:
$\lim_{n \to \infty} \left| \frac{b_{n+1}}{b_n} \right| = \lim_{n \to \infty} \left( \frac{1}{2(2n+1)} \left(1 + \frac{1}{n}\right)^n \right)$
$= \left( \lim_{n \to \infty} \frac{1}{2(2n+1)} \right) \cdot \left( \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \right)$
$= 0 \cdot e = 0$.
- Since the limit is $0 < 1$, series (Q) converges by the Ratio Test.
Conclusion
Series (P) diverges and series (Q) converges.
Important Questions from Infinite Series
- Consider the two series, $S_A$ and $S_B$, where
$$S_A = \sum_{n=1}^\infty \frac{n^2}{2^n}$$
$$S_B = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64} + \frac{1}{128} + \frac{1}{512} + \cdots$$
Which of the following statements is correct for the two given series? - The value of $\sum_{i=0}^{\infty} \sum_{j=1}^{\infty} 2^{-i} 3^{-j}$ is ______________ . (Answer in integer)
- Match each entry of List-1 with a suitable entry in List-2 and choose the correct option.
List-1 List-2 P The sum of the series $\sum_{n=1}^\infty \frac{1}{(n+2)(n+1)}$ is equal to I $\frac{3}{2}$ Q $\lim_{x \to 0} \left( \frac{3}{x^2} \int_0^x \sin(t) dt \right)$ is equal to II $1$ R Let $\frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx)$ be the Fourier series expansion of the function $f(x) = \frac{1}{2} \sin x - \frac{1}{2} \cos x + \frac{1}{\sqrt{2}} \sin 2x, x \in [0, 2\pi]$. Then, $\sum_{n=0}^\infty (a_n^2 + b_n^2)$ is equal to III $\frac{1}{2}$ The sum of the following infinite series is
$2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{8} + \frac{1}{9} + \frac{1}{16} + \frac{1}{27} + \dots$Consider the following two series
P: $\sum_{n=1}^{\infty} \frac{1}{n}$
Q: $\sum_{n=1}^{\infty} \frac{1}{n^2}$
Choose the correct option from the following
