To determine whether the series

$\sum_{n=3}^\infty \frac{\ln n}{(n-2)^2}$

converges or diverges, we can use comparison tests or evaluate the behavior of the terms as $n \to \infty$. Here's the detailed reasoning:

Step 1: Analyze the terms

The term of the series is:

$a_n = \frac{\ln n}{(n-2)^2}.$

As $n \to \infty$, note:

• The numerator $\ln n$ grows logarithmically (much slower than any polynomial).
• The denominator $(n-2)^2$ grows quadratically.

Thus, $a_n$ decreases as $n \to \infty$.

Step 2: Use the Comparison Test

To check for convergence, compare $a_n$ with a simpler, known series. For large $n$, $(n-2)^2$ behaves similarly to $n^2$, so $a_n$ behaves like:

$\frac{\ln n}{n^2}.$

Simplified Comparison

Compare $a_n$ with the series $\sum \frac{\ln n}{n^2}$. The convergence of this comparison series can be tested using the Integral Test.

Step 3: Apply the Integral Test to $\int \frac{\ln x}{x^2} dx$

Let $u = \ln x$, so $du = \frac{1}{x} dx$. Then:

$\int \frac{\ln x}{x^2} dx = \int u e^{-u} du.$

The integral $\int u e^{-u} du$ converges (this is a standard integral, which evaluates to $-e^{-u}(u+1)$). Thus, the integral converges, and so does the series $\sum \frac{\ln n}{n^2}$.

Step 4: Conclude for the Original Series

Since $a_n = \frac{\ln n}{(n-2)^2}$ behaves similarly to $\frac{\ln n}{n^2}$ for large $n$, and $\sum \frac{\ln n}{n^2}$ converges, the original series

$\sum_{n=3}^\infty \frac{\ln n}{(n-2)^2}$

also converges by the Comparison Test.

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Would you like details on any part of this solution or a different approach? Here are five related questions to explore:

1. What is the Comparison Test, and how is it applied to determine convergence?
2. How does the Integral Test work, and when should it be used?
3. Why does $\ln n$ grow slower than any polynomial function of $n$?
4. Can the Ratio Test be used to determine the convergence of this series?
5. What are some examples of divergent series for comparison?

Tip: Always compare the given series with a simpler, well-known series to identify convergence behavior efficiently.