Series Convergence Analysis

We need to determine the convergence of three infinite series:

• (i) $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$
• (ii) $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$
• (iii) $\sum_{n=1}^{\infty} \frac{1}{n!}$

Analysis of Series (i)

The series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ can be written as $\sum_{n=1}^{\infty} \frac{1}{n^{1/2}}$.

This is a p-series of the form $\sum_{n=1}^{\infty} \frac{1}{n^p}$ with $p = 1/2$.

Rule: A p-series converges if $p > 1$ and diverges if $p \le 1$.

Since $p = 1/2 \le 1$, series (i) diverges.

Analysis of Series (ii)

The series is $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$.

We can use partial fraction decomposition: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.

This forms a telescoping series. Let's look at the N-th partial sum, $S_N$:

$S_N = \sum_{n=1}^{N} \left( \frac{1}{n} - \frac{1}{n+1} \right)$

$S_N = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{N} - \frac{1}{N+1}\right)$

After cancellation, $S_N = 1 - \frac{1}{N+1}$.

To find the sum of the infinite series, we take the limit as $N \to \infty$:

$\lim_{N \to \infty} S_N = \lim_{N \to \infty} \left( 1 - \frac{1}{N+1} \right) = 1 - 0 = 1$

Since the limit exists and is finite, series (ii) converges.

Analysis of Series (iii)

The series is $\sum_{n=1}^{\infty} \frac{1}{n!}$.

We can use the Ratio Test. Let $a_n = \frac{1}{n!}$.

Ratio Test: Calculate $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$.

$\lim_{n \to \infty} \left| \frac{1/(n+1)!}{1/n!} \right| = \lim_{n \to \infty} \frac{n!}{(n+1)!} = \lim_{n \to \infty} \frac{n!}{(n+1)n!} = \lim_{n \to \infty} \frac{1}{n+1}$

$\lim_{n \to \infty} \frac{1}{n+1} = 0$

Rule: If the limit is less than 1, the series converges.

Since the limit is $0 < 1$, series (iii) converges.

Conclusion

Based on the analysis:

• Series (i) diverges.
• Series (ii) converges.
• Series (iii) converges.

Therefore, only series (ii) and (iii) converge.