Question

Let \( \{a_n\}_{n \geq 1} \) be a sequence of real numbers such that \[ a_n = \sum_{k=n+1}^{2n} \frac{1}{k}, n \geq 1. \] Then which of the following statement(s) is (are) true?

• A. \( \{a_n\}_{n \geq 1} \) is an increasing sequence
• B. \( \{a_n\}_{n \geq 1} \) is bounded below
• C. \( \{a_n\}_{n \geq 1} \) is bounded above
• D. \( \{a_n\}_{n \geq 1} \) is a convergent sequence

Hint:

For sums of fractions, the behavior of the sequence can often be determined by examining how the upper and lower bounds of the summation change with \( n \).

Answer 1

The Correct Option is A, B, C, D

Step 1: Compute first few terms

$$a_1 = \frac{1}{2}$$ $$a_2 = \frac{1}{3} + \frac{1}{4} = \frac{7}{12} \approx 0.583$$ $$a_3 = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} = \frac{37}{60} \approx 0.617$$

The sequence appears to be increasing.

Option (A): ${a_n}_{n\geq 1}$ is an increasing sequence

Compare $a_{n+1}$ and $a_n$: $$a_{n+1} = \sum_{k=n+2}^{2n+2} \frac{1}{k} = \sum_{k=n+1}^{2n} \frac{1}{k} - \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2}$$

$$a_{n+1} - a_n = -\frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2}$$

$$= \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1}$$

$$= \frac{1}{2n+1} + \frac{1}{2(n+1)} - \frac{1}{n+1} = \frac{1}{2n+1} - \frac{1}{2(n+1)}$$

$$= \frac{2(n+1) - (2n+1)}{(2n+1) \cdot 2(n+1)} = \frac{1}{2(2n+1)(n+1)} > 0$$

So $a_{n+1} > a_n$, making the sequence strictly increasing.

Option (A) is TRUE 

Option (B): ${a_n}_{n\geq 1}$ is bounded below

Since all terms are positive and $a_1 = \frac{1}{2}$, we have $a_n > 0$ for all $n$.

Option (B) is TRUE 

Option (C): ${a_n}_{n\geq 1}$ is bounded above

Using integral approximation: $$a_n = \sum_{k=n+1}^{2n} \frac{1}{k} \approx \int_n^{2n} \frac{1}{x} dx = \ln(2n) - \ln(n) = \ln 2$$

More precisely, for the harmonic series partial sums: $$\sum_{k=n+1}^{2n} \frac{1}{k} < \int_n^{2n} \frac{1}{x} dx = \ln 2$$

And: $$\sum_{k=n+1}^{2n} \frac{1}{k} > \int_{n+1}^{2n+1} \frac{1}{x} dx \to \ln 2 \text{ as } n \to \infty$$

Since $a_n < \ln 2$ for all $n$, the sequence is bounded above.

Option (C) is TRUE 

Option (D): ${a_n}_{n\geq 1}$ is a convergent sequence

Since the sequence is increasing and bounded above (by $\ln 2$), by the Monotone Convergence Theorem, it converges.

In fact, $\lim_{n \to \infty} a_n = \ln 2$.

Option (D) is TRUE 

Answer: (A), (B), (C), and (D)