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The given series is:

\[
\sum_{n=1}^{\infty} \ln \left( \frac{n}{n+1} \right)
\]

To determine if the series is convergent or divergent, let's first simplify the general term:

\[
\ln \left( \frac{n}{n+1} \right) = \ln(n) - \ln(n+1)
\]

This is a **telescoping series**, where consecutive terms cancel out much of the series. Specifically, the series looks like:

\[
\sum_{n=1}^{\infty} \left( \ln(n) - \ln(n+1) \right)
\]

Notice that each term cancels out with the subsequent one, leaving:

\[
\lim_{N \to \infty} \left( \ln(1) - \ln(N+1) \right) = 0 - \infty = -\infty
\]

Thus, the series **diverges** because as \( N \to \infty \), the sum grows without bound in the negative direction.

### Final answer: The series **diverges**.

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Do you have any questions or want further details on the explanation? Here are 5 related questions for deeper understanding:

1. What is a telescoping series, and how does it work?
2. Can a logarithmic series ever converge, and under what conditions?
3. How can you formally prove divergence using other tests, such as the Comparison Test?
4. What happens if the argument of the logarithm were different, say \( \ln\left( \frac{n+1}{n} \right) \)?
5. What are the differences between conditional and absolute convergence in series?

**Tip:** Always check for telescoping patterns in series involving logarithms or fractions to simplify the process of summation.

Determine whether the series is convergent or divergent. If convergent, find the sum; if divergent, enter div. Series: sum from n=1 to infinity of ln(n/(n+1)).

Explore whether the series sum ln(n/(n+1)) is convergent or divergent using telescoping series techniques and logarithmic properties. This problem is ideal for calculus students at the college level.

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