To determine whether the infinite series
$\sum_{k=1}^\infty \frac{\sqrt{k}}{\sqrt{k^2 + 1}}$
converges or diverges using the Divergence Test, let's proceed as follows:

Divergence Test:

The divergence test states that if $\lim_{k \to \infty} a_k \neq 0$, where $a_k$ is the general term of the series, then the series diverges. However, if $\lim_{k \to \infty} a_k = 0$, the test is inconclusive, and other methods must be used.

Here, $a_k = \frac{\sqrt{k}}{\sqrt{k^2 + 1}}$.

Step 1: Compute $\lim_{k \to \infty} a_k$.

$a_k = \frac{\sqrt{k}}{\sqrt{k^2 + 1}}$ Divide numerator and denominator by $\sqrt{k^2}$: $a_k = \frac{\sqrt{k}}{\sqrt{k^2 + 1}} = \frac{\sqrt{k}}{\sqrt{k^2(1 + \frac{1}{k^2})}} = \frac{1}{\sqrt{1 + \frac{1}{k^2}}}.$

Now, as $k \to \infty$, $\frac{1}{k^2} \to 0$. So, $\lim_{k \to \infty} a_k = \frac{1}{\sqrt{1 + 0}} = 1.$

Step 2: Conclusion

Since $\lim_{k \to \infty} a_k = 1 \neq 0$, the series diverges by the Divergence Test.

Would you like additional details or explanations? Here are some questions to deepen understanding:

1. What is the difference between the divergence test and other tests for convergence?
2. Can the divergence test ever confirm that a series converges?
3. How would you approach this series using the Limit Comparison Test?
4. What are some examples of series that converge despite having terms that approach zero?
5. How does the behavior of the general term $a_k$ affect the outcome of the divergence test?

Tip: Always check the behavior of the general term at infinity before applying more advanced convergence tests.