To determine whether each series converges or diverges, we will apply appropriate convergence tests. The most common tools include the comparison test, limit comparison test, ratio test, root test, and knowledge of well-known convergent/divergent series like the p-series or geometric series. Here’s the analysis for each:

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(a) $\sum_{k=2}^{\infty} \frac{1}{k^{\sqrt{2}} - 1}$

Test Used: Comparison Test

Let's compare this series to a simpler series. As $k \to \infty$, the term $k^{\sqrt{2}} - 1$ behaves very similarly to $k^{\sqrt{2}}$, since the "-1" becomes insignificant for large $k$. Thus, for large $k$, the term $\frac{1}{k^{\sqrt{2}} - 1}$ behaves like $\frac{1}{k^{\sqrt{2}}}$.

Now consider the series $\sum \frac{1}{k^{\sqrt{2}}}$, which is a p-series with $p = \sqrt{2}$. Since $\sqrt{2} > 1$, this p-series converges.

By comparison, $\sum \frac{1}{k^{\sqrt{2}} - 1}$ converges as well.

Conclusion: Converges

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(b) $\sum_{k=1}^{\infty} \frac{r^k}{k^3 + 1}$ (where $0 \leq r < 1$)

Test Used: Ratio Test (and knowledge of geometric series)

This series contains a geometric component $r^k$ and a decaying term $\frac{1}{k^3 + 1}$.

1. For large $k$, $k^3 + 1 \sim k^3$, so $\frac{1}{k^3 + 1}$ behaves like $\frac{1}{k^3}$.
2. Consider just the geometric component: $r^k$ decays exponentially since $0 \leq r < 1$.
3. The $k^3$ term in the denominator decays polynomially.

Since $r^k$ decays exponentially, this decay dominates the polynomial growth of $k^3 + 1$, and the series converges by the ratio test.

Conclusion: Converges

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(c) $\sum_{k=0}^{\infty} \frac{2^{k+2} + k^2}{3^k + k + 12}$

Test Used: Dominant Term Analysis (Asymptotic Behavior)

For large $k$, the dominant terms in the numerator and denominator are:

• Numerator: $2^{k+2} \sim 2^k$
• Denominator: $3^k$

Thus, for large $k$, the terms in the series behave like $\frac{2^k}{3^k}$.

Simplifying this expression: $\frac{2^k}{3^k} = \left(\frac{2}{3}\right)^k$

This is a geometric series with a ratio $\frac{2}{3}$, which is less than 1. A geometric series with $|r| < 1$ converges.

Conclusion: Converges

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(d) $\sum_{k=3}^{\infty} \frac{1}{k^2 \ln(k)}$

Test Used: Integral Test

We use the integral test here, since this series involves a function that is continuous, positive, and decreasing for large $k$.

The corresponding integral to evaluate is: $\int_{3}^{\infty} \frac{1}{x^2 \ln(x)} dx$

Using integration by substitution (with $u = \ln(x)$, so $du = \frac{1}{x} dx$), we find that the integral converges. Therefore, by the integral test, the series also converges.

Conclusion: Converges

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Summary:

• (a) Converges
• (b) Converges
• (c) Converges
• (d) Converges

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Would you like any details or further explanations on any of the tests used? Here are five related questions:

1. How does the comparison test work, and when is it applicable?
2. What is the p-series test, and how do we determine convergence for different values of $p$?
3. Can the ratio test be used for series with more complicated terms? How do we apply it?
4. What is the integral test, and why does it work for some series but not others?
5. How can asymptotic analysis help in determining the behavior of terms in a series?

Tip: When using the ratio test, always check the limit of the ratio as $k \to \infty$, and remember that the test works well when the series behaves like a geometric series.