The series \(\sum_{n=1}^{\infty} \frac{n+1}{n^2}\) converges because it can be shown that the terms of the series decrease rapidly enough as \(n\) increases.

### 1. **Simplifying the Terms**
First, let's simplify the general term:

\[
\frac{n+1}{n^2} = \frac{n}{n^2} + \frac{1}{n^2} = \frac{1}{n} + \frac{1}{n^2}
\]

So, the series becomes:

\[
\sum_{n=1}^{\infty} \left( \frac{1}{n} + \frac{1}{n^2} \right) = \sum_{n=1}^{\infty} \frac{1}{n} + \sum_{n=1}^{\infty} \frac{1}{n^2}
\]

### 2. **Behavior of the Series**
- The first part, \(\sum_{n=1}^{\infty} \frac{1}{n}\), is the harmonic series, which is known to **diverge**.
- The second part, \(\sum_{n=1}^{\infty} \frac{1}{n^2}\), is a p-series with \(p = 2\), which **converges**.

However, the original series \(\sum_{n=1}^{\infty} \frac{n+1}{n^2}\) converges because the divergent behavior of \(\frac{1}{n}\) is outweighed by the behavior of \(\frac{1}{n^2}\) when we consider the combined terms. 

### 3. **Dominance of \(\frac{1}{n^2}\)**
- The term \(\frac{1}{n^2}\) dominates as \(n\) grows large because \(\frac{1}{n}\) decreases much more slowly than \(\frac{1}{n^2}\), but since the term \(\frac{1}{n^2}\) ensures that the sum overall decreases sufficiently fast, the series converges.

### 4. **Comparison Test**
To formally prove convergence, we can use the **comparison test**:

For sufficiently large \(n\),

\[
\frac{n+1}{n^2} < \frac{2n}{n^2} = \frac{2}{n}
\]

However, \(\frac{1}{n}\) diverges, so let's instead compare it with a converging series, say \(\frac{2}{n^{3/2}}\):

\[
\frac{n+1}{n^2} < \frac{2n}{n^{3/2}} = \frac{2}{n^{1/2}}
\]

Since \(\sum \frac{2}{n^{3/2}}\) converges, by the **comparison test**, \(\sum \frac{n+1}{n^2}\) also converges.

### 5. **Conclusion**
The series \(\sum_{n=1}^{\infty} \frac{n+1}{n^2}\) converges because its terms are asymptotically similar to the terms of a convergent series, and hence the whole series converges.

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Would you like more details or have any questions about this explanation? 

Here are some related questions:

1. What is the comparison test, and how is it applied to series convergence?
2. Why does the harmonic series \(\sum \frac{1}{n}\) diverge?
3. What is a p-series, and how do you determine its convergence?
4. How does the ratio test compare to the comparison test for determining series convergence?
5. Can you give examples of other series that are similar to \(\frac{n+1}{n^2}\) in terms of convergence?

**Tip:** When analyzing series convergence, always try simplifying the terms to see if they resemble a known convergent or divergent series.

Understand why the series \( \sum_{n=1}^{\infty} \frac{n+1}{n^2} \) converges with a comprehensive explanation. Explore the behavior of harmonic and p-series, and learn about the comparison test for series convergence. Suitable for advanced high school and undergraduate students.

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