Question

Let {a_n}_n≥1 be a sequence of real numbers such that \(a_n=\frac{1}{3^n}\) for all n ≥ 1. Then which of the following statements is/are true ?

• A. \(\sum^{\infin}_{n=1}(-1)^{n+1}a_n\) is a convergent series
• B. \(\sum^{\infin}_{n=1}\frac{(-1)^{n+1}}{n}(a_1+a_2+...+a_n)\) is a convergent series
• C. The radius of convergence of the power series \(\sum^{\infin}_{n=1}a_nx^n \text{ is }\frac{1}{3}\)
• D. \(\sum^{\infin}_{n=1}a_n\sin\frac{1}{a_n}\) is a convergent series

Answer 1

The Correct Option is A, B, D

Let us analyze the given sequence and statements one by one:

The sequence provided is \( a_n = \frac{1}{3^n} \) for all \( n \geq 1 \).

1. Statement 1: \( \sum^{\infty}_{n=1}(-1)^{n+1}a_n \) is a convergent series.

   This series is an alternating series since the terms alternate in sign due to \( (-1)^{n+1} \). By the Alternating Series Test (Leibniz Test), an alternating series \( \sum (-1)^n b_n \) converges if:

   • The sequence \( b_n \) decreases monotonically: \( b_{n+1} \leq b_n \)
   • \( \lim_{n \to \infty} b_n = 0 \)

   Here, \( b_n = a_n = \frac{1}{3^n} \). Clearly, \( b_n \) is decreasing and approaches 0 as \( n \to \infty \). Therefore, this series converges.

2. Statement 2: \( \sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n}(a_1+a_2+\ldots+a_n) \) is a convergent series.

   The series inside the summation, \( a_1 + a_2 + \ldots + a_n \), is a geometric series with terms \( \frac{1}{3^1} + \frac{1}{3^2} + \ldots + \frac{1}{3^n} \). Its sum \( S_n = \frac{1/3(1-(1/3)^n)}{1 - 1/3} = \frac{1}{2}(1 - (1/3)^n) \), approaching \(\frac{1}{2}\) as \( n \to \infty \).

   The outer series \( \sum \frac{(-1)^{n+1}}{n} S_n \) is an alternating series with terms tending to 0. Hence, it converges by the alternating series test.

3. Statement 3: The radius of convergence of the power series \( \sum^{\infty}_{n=1} a_n x^n \text{ is }\frac{1}{3}\).

   For a power series \( \sum a_n x^n \), the radius of convergence \( R \) is found using the formula:

   \( \frac{1}{R} = \limsup_{n \to \infty} \left| a_n \right|^{1/n} \)

   Here \( a_n = \frac{1}{3^n} \), so:

   \( \left| a_n \right|^{1/n} = \left( \frac{1}{3^n} \right)^{1/n} = \frac{1}{3} \)

   The radius of convergence \( R = 3 \), not \( \frac{1}{3} \). Hence, this statement is incorrect.

4. Statement 4: \( \sum^{\infty}_{n=1} a_n \sin\frac{1}{a_n} \) is a convergent series.

   Since \( a_n = \frac{1}{3^n} \), we have:

   \( \sin\left(\frac{1}{a_n}\right) = \sin(3^n) \). Since \( \sin(x) \) is bounded between -1 and 1, the terms of the series can be simplified as:

   \( a_n \sin\left(\frac{1}{a_n}\right) \leq \left|\frac{1}{3^n}\right| \)

   The convergence of \( \sum \frac{1}{3^n} \) (a geometric series with ratio \( \frac{1}{3} \)) implies the convergence of \( \sum a_n \sin(\frac{1}{a_n}) \).

Conclusion: The correct statements are:

• \( \sum^{\infty}_{n=1}(-1)^{n+1}a_n \) is a convergent series.
• \( \sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n}(a_1+a_2+...+a_n) \) is a convergent series.
• \( \sum^{\infty}_{n=1}a_n\sin\frac{1}{a_n} \) is a convergent series.