Step-by-step telescoping summation:
Each term can be decomposed into partial fractions:

$$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$$

Writing out the sum for $k = 1$ to $9$:

$$S = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{9} - \frac{1}{10}\right)$$

All intermediate terms cancel out:

$$S = 1 - \frac{1}{10} = \frac{9}{10}$$

• Incorrect: $10/9$ is the reciprocal.
• Correct: Accurately equals $9/10$.
• Incorrect: $5/6$ would be the sum up to 5 terms.
• Incorrect: $6/5$ is invalid.

The series $\sum_{n=1}^\infty \frac{1}{n}$ is the standard harmonic series.
By the p-series test, $\sum_{n=1}^\infty \frac{1}{n^p}$ converges if $p \gt 1$ and diverges if $p \le 1$.
Here, $p = 1$, so the series diverges to infinity.

Detailed analysis of options:

• Convergent is incorrect because the partial sums grow without bound (as $\ln n$).
• Divergent is correct; the harmonic series is a classic divergent series.
• Absolutely convergent is incorrect as it is not convergent.
• Conditionally convergent applies to alternating series, not positive term divergent series.

A fundamental theorem of real analysis states: 'Every convergent sequence of real numbers is bounded.' Thus, boundedness is a necessary condition for convergence (though not sufficient, e.g., $(-1)^n$ is bounded but not convergent).

Detailed analysis of options:

• Bounded is the correct necessary condition; an unbounded sequence can never converge.
• Having more than one limit is impossible because the limit of a convergent sequence is unique.
• Monotonic is not necessary; for example, $a_n = \frac{(-1)^n}{n}$ converges to 0 but oscillates and is not monotonic.
• Not Cauchy is incorrect because every convergent sequence in $\mathbb{R}$ must be a Cauchy sequence.