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.

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.

Behavior of real sequences:

• $(a_n) = (-1)^n$ does not converge because it oscillates finitely between $-1$ and $+1$.
• $(b_n) = 2^n$ does not converge because it grows without bound to $+\infty$.
• Neither sequence converges to a finite real limit.
• $(a_n) = (-1)^n$ is an oscillatory sequence with two distinct subsequential limits ($-1$ and $+1$), hence divergent. $(b_n) = 2^n$ tends to $+\infty$, hence divergent. Therefore, both $(a_n)$ and $(b_n)$ are divergent.