Question

Let the series 𝑆 and 𝑇 be defined by
\(∑^∞_{n-0}\frac{ 2 ⋅ 5 ⋅ 8 ⋯ (3𝑛 + 2) }{1 ⋅ 5 ⋅ 9 ⋯ (4𝑛 + 1)}\) and \(∑^∞ _{n=1}(1 + \frac{1 }{𝑛} ) ^{−𝑛^2}\)
respectively. Then, which one of the following statements is TRUE?

• A. 𝑆 is convergent and 𝑇 is divergen
• B. 𝑆 is divergent and 𝑇 is convergent
• C. Both 𝑆 and 𝑇 are convergent
• D. Both 𝑆 and 𝑇 are divergent

Answer 1

The Correct Option is C

The problem requires evaluating the convergence of two infinite series, \(S\) and \(T\), defined as follows:

Series \(S\):

\(S = ∑^∞_{n=0}\frac{2 ⋅ 5 ⋅ 8 ⋯ (3n + 2)}{1 ⋅ 5 ⋅ 9 ⋯ (4n + 1)}\)

Series \(T\):

\(T = ∑^∞_{n=1}(1 + \frac{1}{n})^{-n^2}\)

Let's analyze each series: 

1. Convergence of Series \(S\):
   The series \(S\) can be represented as a product where the numerators and denominators follow an arithmetic pattern:
   • The terms in the numerator are in arithmetic progression with the form \(3n + 2\).
   • The terms in the denominator are in arithmetic progression with the form \(4n + 1\).
2. Convergence of Series \(T\):
   The series is defined as: \(T = ∑^∞_{n=1}(1 + \frac{1}{n})^{-n^2}\) Considering the expression: \((1 + \frac{1}{n})^{-n^2} = e^{-n \cdot \ln(1+\frac{1}{n})}\) For large \(n\), we approximate \(\ln(1+\frac{1}{n}) \approx \frac{1}{n}\). Thus: \(e^{-n \cdot \ln(1+\frac{1}{n})} \approx e^{-n \cdot \frac{1}{n}} = e^{-1}\) Each term of the series can be approximated to a constant that ensures the finite nature of terms, making it convergent.

By analyzing both series, we determine that both \(S\) and \(T\) are convergent.

Conclusion: The correct answer is: Both \(S\) and \(T\) are convergent.