What comparison series would you use with the Comparison Test to determine whether ∑ (k = 1 to ∞) 2ᵏ / (3ᵏ + 1) converges?
Verified step by step guidance
1
Identify the general term of the series: \(a_k = \frac{2^k}{3^k + 1}\).
To apply the Comparison Test, find a simpler series \(b_k\) that resembles \(a_k\) for large \(k\) and whose convergence behavior is known.
Since \$3^k\( dominates the \)+1\( in the denominator for large \)k\(, approximate \)a_k\( by \)\frac{2^k}{3^k}$.
Simplify the approximation: \(\frac{2^k}{3^k} = \left(\frac{2}{3}\right)^k\).
Use the geometric series \(\sum_{k=1}^\infty \left(\frac{2}{3}\right)^k\) as the comparison series, because it converges (common ratio \(\frac{2}{3} < 1\)).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above
Video duration:
2m
Play a video:
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Comparison Test
The Comparison Test determines the convergence or divergence of a series by comparing it to another series with known behavior. If the terms of the given series are smaller than those of a convergent series, it also converges; if larger than those of a divergent series, it diverges.
A geometric series has terms of the form ar^k, where r is the common ratio. It converges if |r| < 1 and diverges otherwise. Recognizing geometric behavior helps in choosing a suitable comparison series for convergence tests.
Analyzing the dominant terms in the numerator and denominator for large k helps simplify the series term. For example, 2^k/(3^k + 1) behaves like (2/3)^k for large k, guiding the choice of a comparison series with similar asymptotic behavior.