We are asked to study the convergence of the series

$\sum_{n=2}^{\infty} \frac{1}{\ln(n^2 + 2)}.$

Step 1: Behavior of the terms for large $n$

For large $n$, the expression $\ln(n^2 + 2)$ behaves similarly to $\ln(n^2)$, since $n^2 + 2$ grows large as $n$ increases. We can approximate $\ln(n^2 + 2)$ as:

$\ln(n^2 + 2) \sim \ln(n^2) = 2\ln(n).$

So, for large $n$, the general term $\frac{1}{\ln(n^2 + 2)}$ behaves like:

$\frac{1}{\ln(n^2 + 2)} \sim \frac{1}{2\ln(n)}.$

Step 2: Comparison with a known series

The series we are considering has terms similar to $\frac{1}{\ln(n)}$, which is related to a known divergent series. To study the convergence more rigorously, we can use the Integral Test or compare the series to a simpler series with a similar form.

• The series $\sum \frac{1}{\ln(n)}$ is known to diverge. This can be shown using the Integral Test by considering the integral:

$\int_{2}^{\infty} \frac{1}{\ln(x)} \, dx,$

which also diverges.

Since for large $n$,

$\frac{1}{\ln(n^2 + 2)} \sim \frac{1}{2\ln(n)},$

the given series behaves similarly to the series $\sum \frac{1}{\ln(n)}$, which is known to diverge.

Step 3: Conclusion

By the Comparison Test (since $\frac{1}{\ln(n^2 + 2)} \sim \frac{1}{2\ln(n)}$), the given series diverges because it behaves similarly to the divergent series $\sum \frac{1}{\ln(n)}$.

Thus, the series

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

diverges.

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Would you like more details or clarification on any part of the reasoning? Here are 5 related questions for further exploration:

1. How does the Integral Test work for series like $\sum \frac{1}{\ln(n)}$?
2. Can you explain the Comparison Test in more depth with examples?
3. How does the divergence of the series $\sum \frac{1}{\ln(n)}$ relate to other common divergent series?
4. Could the series converge if we altered the function, such as $\frac{1}{\ln(n^2 + 2^2)}$?
5. What would happen if the series were $\sum \frac{1}{n^p \ln(n)}$ for some power $p$?

Tip: For series that involve logarithmic functions, it can be helpful to approximate them for large $n$, since these approximations often allow for comparison with simpler series whose convergence properties are well-known.