Determine the convergence or divergence of the series. 
${\sum }_{k=1}^{\infty }\frac{7k^2}{5k+3}$

• A. Inconclusive
• B. Diverges
• C. Converges

Answer: (B)

Verified step by step guidance

1. Step 1: Recognize that the problem involves determining the convergence or divergence of the infinite series \( \sum_{k=1}^{\infty} \frac{7k^2}{5k+3} \). To analyze this, we can use standard tests for convergence such as the Comparison Test, Limit Comparison Test, or the Ratio Test.
2. Step 2: Observe the general term \( \frac{7k^2}{5k+3} \). As \( k \to \infty \), the dominant term in the numerator is \( 7k^2 \) and in the denominator is \( 5k \). Simplify the fraction for large \( k \) to approximate \( \frac{7k^2}{5k+3} \approx \frac{7k^2}{5k} = \frac{7k}{5} \). This suggests the series behaves similarly to \( \sum_{k=1}^{\infty} k \).
3. Step 3: Recall that the series \( \sum_{k=1}^{\infty} k \) is a divergent series because the terms do not approach zero and the sum grows without bound. This provides an initial indication that \( \sum_{k=1}^{\infty} \frac{7k^2}{5k+3} \) may also diverge.
4. Step 4: Apply the Comparison Test or Limit Comparison Test to confirm divergence. Compare \( \frac{7k^2}{5k+3} \) with \( k \). Compute the limit \( \lim_{k \to \infty} \frac{\frac{7k^2}{5k+3}}{k} \). Simplify the expression to find \( \lim_{k \to \infty} \frac{7k^2}{k(5k+3)} = \lim_{k \to \infty} \frac{7k}{5k+3} \). As \( k \to \infty \), this limit approaches \( \frac{7}{5} \), which is a positive finite constant.
5. Step 5: Conclude that since \( \frac{7k^2}{5k+3} \) behaves similarly to \( k \) and \( \sum_{k=1}^{\infty} k \) diverges, the original series \( \sum_{k=1}^{\infty} \frac{7k^2}{5k+3} \) also diverges by the Limit Comparison Test.