Step 1: Understanding the Concept:
We are given a sequence defined by a sum and need to determine its convergence properties. A common method for sequences defined by sums is to use the Squeeze (or Sandwich) Theorem by finding lower and upper bounds for the terms in the sum.
Step 2: Key Formula or Approach:
The Squeeze Theorem: If \(a_n \le b_n \le c_n\) for all \(n\) beyond some point, and \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L\), then \(\lim_{n \to \infty} b_n = L\).
We will bound the sum \(S_n = \sum_{k=1}^n \frac{n}{n^2+k}\).
For \(1 \le k \le n\), the denominator \(n^2+k\) satisfies:
\[ n^2+1 \le n^2+k \le n^2+n \]
Taking the reciprocal reverses the inequalities:
\[ \frac{1}{n^2+n} \le \frac{1}{n^2+k} \le \frac{1}{n^2+1} \]
Multiplying by n:
\[ \frac{n}{n^2+n} \le \frac{n}{n^2+k} \le \frac{n}{n^2+1} \]
Step 3: Detailed Explanation:
We have the inequality for the general term:
\[ \frac{n}{n^2+n} \le \frac{n}{n^2+k} \le \frac{n}{n^2+1} \]
Now, we sum from \(k=1\) to \(n\). The lower and upper bounds do not depend on \(k\), so we are summing them \(n\) times:
\[ \sum_{k=1}^n \frac{n}{n^2+n} \le \sum_{k=1}^n \frac{n}{n^2+k} \le \sum_{k=1}^n \frac{n}{n^2+1} \]
\[ n \cdot \frac{n}{n^2+n} \le S_n \le n \cdot \frac{n}{n^2+1} \]
\[ \frac{n^2}{n(n+1)} \le S_n \le \frac{n^2}{n^2+1} \]
\[ \frac{n}{n+1} \le S_n \le \frac{n^2}{n^2+1} \]
Now, we find the limits of the lower and upper bounds as \(n \to \infty\):
\[ \lim_{n \to \infty} \frac{n}{n+1} = \lim_{n \to \infty} \frac{1}{1+1/n} = \frac{1}{1+0} = 1 \]
\[ \lim_{n \to \infty} \frac{n^2}{n^2+1} = \lim_{n \to \infty} \frac{1}{1+1/n^2} = \frac{1}{1+0} = 1 \]
Since \(S_n\) is squeezed between two sequences that both converge to 1, by the Squeeze Theorem, the sequence \(\{S_n\}\) must also converge to 1.
Therefore, the sequence is convergent.
Note on Discrepancy:
The calculation shows the sequence converges. The provided answer key marks (C) "bounded but not convergent". This is a contradiction. There may be a typo in the question. Let's re-read the OCR. It says \(\frac{n}{n^2+k}\). The sum is \(\sum_{k=1}^n \frac{n}{n^2+k}\). The calculation is correct.
Let's consider another common form: \(S_n = \sum_{k=1}^n \frac{1}{n+k}\). This can be seen as a Riemann sum for \(\int_0^1 \frac{1}{1+x} dx = [\ln(1+x)]_0^1 = \ln(2)\). So this form converges.
Let's re-examine the OCR: \(S_n=\sum_{k=1}^n \frac{n}{n^2+k}\). The work above is correct for this expression. The sequence is convergent. The provided answer key is incorrect. The question might have been \(S_n = \sum_{k=1}^n \frac{n}{n+k^2}\) or something else entirely. Given the provided question, the answer should be (A). I will proceed by stating the correct mathematical conclusion.
Step 4: Final Answer:
By the Squeeze Theorem, the sequence \(\{S_n\}\) converges to 1. Thus, the correct option should be (A) Convergent. The provided answer key appears to be incorrect for the question as stated.