Question

Consider the following series: 
(i) \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \) 
(ii) \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \) 
(iii) \( \sum_{n=1}^{\infty} \frac{1}{n!} \) 
Choose the correct option.

• A. Only (ii) converges
• B. Only (ii) and (iii) converge
• C. Only (iii) converges
• D. All three converge

Hint:

To determine the convergence of a series, recognize if it is a p-series or check for special forms like telescoping series or factorial terms.

Answer 1

The Correct Option is B

Let's examine the convergence of each series: 
(i) \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \): 
This is a p-series with \( p = \frac{1}{2} \), and we know that a p-series converges if \( p > 1 \) and diverges if \( p \leq 1 \). Since \( p = \frac{1}{2} \), this series diverges. 
(ii) \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \): 
We can decompose this into partial fractions: \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}. \] This gives us a telescoping series, where most terms cancel out. The sum of the series converges, so this series converges. 
(iii) \( \sum_{n=1}^{\infty} \frac{1}{n!} \): 
The factorial function grows extremely fast, and it is known that the series \( \sum_{n=1}^{\infty} \frac{1}{n!} \) converges to \( e - 1 \), so this series converges. 
Step 2: Conclusion. Since series (ii) and (iii) converge and series (i) diverges, the correct answer is (B).