Question

Let R₁ and R₂ be the radii of convergence of the power series \(\sum\limits_{n=1}^{\infin}(-1)^nx^{n-1}\) and \(\sum\limits_{n=1}^{\infin}(-1)^n\frac{x^{n+1}}{n(n+1)}\), respectively. Then

• A. R₁ = R₂
• B. R₂ > 1
• C. \(\sum\limits_{n=1}^{\infin}(-1)^nx^{n-1}\) converges for all x ∈ [−1, 1]
• D. \(\sum\limits_{n=1}^{\infin}(-1)^n\frac{x^{n+1}}{n(n+1)}\) converges for all x ∈ [−1, 1]

Answer 1

The Correct Option is A, D

To find the radii of convergence \(R_1\) and \(R_2\) of the given power series, we use the formula for the radius of convergence of a power series \(\sum a_n x^n\), which is given by:

\(R = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}}\) 

Let's analyze each series:

For the series \(S_1 = \sum\limits_{n=1}^{\infty} (-1)^n x^{n-1}\), rewrite it as:

\(S_1 = \sum\limits_{n=0}^{\infty} (-1)^{n+1} x^n\)

Here, the terms are \(a_n = (-1)^{n+1}\).

Thus, \(|a_n| = 1\) for all \(n\), so the limsup is 1. Hence, the radius of convergence \(R_1\) is:

\(R_1 = \frac{1}{1} = 1\)

For the series \(S_2 = \sum\limits_{n=1}^{\infty} (-1)^n \frac{x^{n+1}}{n(n+1)}\), rewrite it as:

\(S_2 = \sum\limits_{n=1}^{\infty} a_n x^{n+1}\), where \(a_n = \frac{(-1)^n}{n(n+1)}\)

Now, \(|a_n| = \frac{1}{n(n+1)}\),

\(\limsup_{n \to \infty} \sqrt[n]{|a_n|} = \limsup_{n \to \infty} \left(\frac{1}{n(n+1)}\right)^{1/n} = 1\) because \(|a_n|\) tends to 0 as \(n\) approaches infinity.

Thus, the radius of convergence \(R_2\) is:

\(R_2 = \frac{1}{1} = 1\)

Therefore, \(R_1 = R_2 = 1\). This means both series converge for \(|x| < 1\). We need to check their convergence at the endpoints \(-1\) and \(1\).

At \(x = -1\), \(S_1\) becomes the alternating harmonic series:

\(S_1 = \sum\limits_{n=0}^{\infty} (-1)^{n+1} (-1)^n = \sum\limits_{n=0}^{\infty} (-1)^{2n+1} = \sum\limits_{n=0}^{\infty} (-1)^{n+1}\)

This series converges by the Alternating Series Test.

Similarly, at \(x = -1\), \(S_2\) becomes:

\(S_2 = \sum\limits_{n=1}^{\infty} (-1)^n \frac{(-1)^{n+1}}{n(n+1)} = \sum\limits_{n=1}^{\infty} \frac{1}{n(n+1)}\)

This series is telescoping and converges.

Thus, option \(R_1 = R_2\) is correct. Also, since \(S_2\) converges for all \(x \in [-1, 1]\), the correct answers are:

\(R_1 = R_2\)

,

\(\sum\limits_{n=1}^{\infin}(-1)^n\frac{x^{n+1}}{n(n+1)}\) converges for all \(x \in [-1, 1]\)