Math worksheet featuring series convergence problems using comparison tests.

To solve the problems using the Comparison Test or Limit Comparison Test, we need to determine whether the given series converge or diverge by comparing them to known convergent or divergent series. Let's go through each problem step by step.

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Problem 27:

$$
\sum_{k=1}^\infty \frac{1}{k^2 + 4}
$$

#### Solution:
1. Compare with a known series:
The term $\frac{1}{k^2 + 4}$ is similar to $\frac{1}{k^2}$ for large $k$, since the $+4$ in the denominator becomes negligible as $k \to \infty$.

2. Use the Comparison Test:
For all $k \geq 1$,
$$
\frac{1}{k^2 + 4} < \frac{1}{k^2}.
$$
The series $\sum_{k=1}^\infty \frac{1}{k^2}$ is a convergent $p$-series with $p = 2 > 1$.

3. Conclusion:
Since $\frac{1}{k^2 + 4} < \frac{1}{k^2}$ and $\sum_{k=1}^\infty \frac{1}{k^2}$ converges, by the Comparison Test, the series $\sum_{k=1}^\infty \frac{1}{k^2 + 4}$ also converges.

$$
\boxed{\text{Converges}}
$$

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Problem 30:

$$
\sum_{k=1}^\infty \frac{0.0001}{k + 4}
$$

#### Solution:
1. Simplify the term:
The term $\frac{0.0001}{k + 4}$ can be compared to $\frac{0.0001}{k}$ for large $k$, since the $+4$ in the denominator becomes negligible as $k \to \infty$.

2. Use the Comparison Test:
For all $k \geq 1$,
$$
\frac{0.0001}{k + 4} < \frac{0.0001}{k}.
$$
The series $\sum_{k=1}^\infty \frac{0.0001}{k}$ is a constant multiple of the harmonic series $\sum_{k=1}^\infty \frac{1}{k}$, which diverges.

3. Conclusion:
Since $\frac{0.0001}{k + 4} \sim \frac{0.0001}{k}$ and $\sum_{k=1}^\infty \frac{0.0001}{k}$ diverges, by the Comparison Test, the series $\sum_{k=1}^\infty \frac{0.0001}{k + 4}$ also diverges.

$$
\boxed{\text{Diverges}}
$$

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Problem 33:

$$
\sum_{k=1}^\infty \frac{\sin(1/k)}{k^2}
$$

#### Solution:
1. Behavior of $\sin(1/k)$:
For small $x$, $\sin(x) \approx x$. Therefore, for large $k$, $\sin(1/k) \approx \frac{1}{k}$.

2. Approximate the term:
For large $k$,
$$
\frac{\sin(1/k)}{k^2} \approx \frac{1/k}{k^2} = \frac{1}{k^3}.
$$

3. Use the Limit Comparison Test:
Compare $\frac{\sin(1/k)}{k^2}$ with $\frac{1}{k^3}$. Compute the limit:
$$
\lim_{k \to \infty} \frac{\frac{\sin(1/k)}{k^2}}{\frac{1}{k^3}} = \lim_{k \to \infty} \frac{\sin(1/k)}{k^2} \cdot k^3 = \lim_{k \to \infty} k \sin(1/k).
$$
Using the fact that $\sin(1/k) \approx \frac{1}{k}$ for large $k$,
$$
\lim_{k \to \infty} k \sin(1/k) = \lim_{k \to \infty} k \cdot \frac{1}{k} = 1.
$$

4. Conclusion:
Since the limit is a positive finite number, the series $\sum_{k=1}^\infty \frac{\sin(1/k)}{k^2}$ behaves like $\sum_{k=1}^\infty \frac{1}{k^3}$, which is a convergent $p$-series with $p = 3 > 1$. Therefore, the series converges.

$$
\boxed{\text{Converges}}
$$

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Problem 38:

$$
\sum_{k=2}^\infty \frac{1}{(k \ln k)^2}
$$

#### Solution:
1. Compare with a known series:
The term $\frac{1}{(k \ln k)^2}$ suggests a comparison with $\frac{1}{k^p}$ for some $p > 1$. However, the presence of $\ln k$ makes it more complex. We use the integral test or comparison with a simpler series.

2. Use the Comparison Test:
For large $k$, $(k \ln k)^2$ grows faster than $k^2$, so $\frac{1}{(k \ln k)^2}$ decreases faster than $\frac{1}{k^2}$. Specifically,
$$
\frac{1}{(k \ln k)^2} < \frac{1}{k^2}.
$$
The series $\sum_{k=2}^\infty \frac{1}{k^2}$ is a convergent $p$-series with $p = 2 > 1$.

3. Conclusion:
Since $\frac{1}{(k \ln k)^2} < \frac{1}{k^2}$ and $\sum_{k=2}^\infty \frac{1}{k^2}$ converges, by the Comparison Test, the series $\sum_{k=2}^\infty \frac{1}{(k \ln k)^2}$ also converges.

$$
\boxed{\text{Converges}}
$$

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Final Answers:

- Problem 27: $\boxed{\text{Converges}}$
- Problem 30: $\boxed{\text{Diverges}}$
- Problem 33: $\boxed{\text{Converges}}$
- Problem 38: $\boxed{\text{Converges}}$