The image contains a problem about determining the convergence or divergence of a series using the comparison test. Let me analyze it step by step and summarize the reasoning behind convergence or divergence.Here’s a breakdown of the given problem and solution:

1. Series Under Analysis

The given series is: $\sum_{k=1}^\infty \frac{2\sqrt{k}}{4k - 3}.$

2. Terms of the Series and Downward Comparison

Using the inequality $4k - 3 < 4k$, we estimate: $\frac{2\sqrt{k}}{4k - 3} > \frac{2\sqrt{k}}{4k}.$ Simplifying the right-hand side: $\frac{2\sqrt{k}}{4k} = \frac{1}{2} \cdot \frac{\sqrt{k}}{k} = \frac{1}{2k^{1/2}}.$ This means the terms are bounded below by a series proportional to $\frac{1}{k^{1/2}}$, which is a divergent $p$-series ($p = 1/2 < 1$).

3. Upward Comparison

Using $4k - 3 > 3k$, we estimate: $\frac{2\sqrt{k}}{4k - 3} < \frac{2\sqrt{k}}{3k}.$ Simplifying: $\frac{2\sqrt{k}}{3k} = \frac{2}{3} \cdot \frac{1}{k^{1/2}}.$ This bounds the terms above by another divergent $p$-series $\frac{1}{k^{1/2}}$.

4. Comparison Test

To apply the comparison test, compare with the series: $\sum_{k=1}^\infty \frac{1}{k^{1/2}}.$ Since both the lower and upper bounds are divergent $p$-series, the original series also diverges by the comparison test.

5. Claim and Conclusion

The series $\sum_{k=1}^\infty \frac{2\sqrt{k}}{4k - 3}$ diverges.

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