prove an = n/(n^2+1) + n/(n^2+2) .... n/(n^2+n) is a convergent sequence using sandwich theorem

To prove that the sequence

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

is convergent using the **sandwich (squeeze) theorem**, we need to find two sequences, say \( b_n \) and \( c_n \), such that:

\[
b_n \leq a_n \leq c_n \quad \text{for all } n
\]

and both \( b_n \) and \( c_n \) converge to the same limit as \( n \to \infty \). Here's how we proceed:

### Step 1: Express the General Term
The given sequence \( a_n \) is a sum of terms of the form

\[
\frac{n}{n^2 + k}, \quad k = 1, 2, \dots, n
\]

Let's analyze the behavior of each individual term. For large \( n \), we approximate:

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

Thus, for large \( n \), each term behaves like \( \frac{1}{n} \).

### Step 2: Find Upper Bound \( c_n \)
Since \( k \geq 1 \), we have:

\[
n^2 + k \geq n^2 + 1
\]

Thus,

\[
\frac{n}{n^2 + k} \leq \frac{n}{n^2 + 1}
\]

Summing this inequality over all \( k \) from 1 to \( n \), we get:

\[
a_n = \sum_{k=1}^{n} \frac{n}{n^2 + k} \leq \sum_{k=1}^{n} \frac{n}{n^2 + 1} = n \cdot \frac{n}{n^2 + 1}
\]

Simplifying the right-hand side:

\[
a_n \leq \frac{n^2}{n^2 + 1}
\]

For large \( n \), \( \frac{n^2}{n^2 + 1} \to 1 \).

Thus, we have an upper bound:

\[
a_n \leq 1
\]

### Step 3: Find Lower Bound \( b_n \)
Similarly, since \( k \leq n \), we have:

\[
n^2 + k \leq n^2 + n
\]

Thus,

\[
\frac{n}{n^2 + k} \geq \frac{n}{n^2 + n} = \frac{n}{n(n + 1)} = \frac{1}{n + 1}
\]

Summing this inequality over all \( k \) from 1 to \( n \), we get:

\[
a_n = \sum_{k=1}^{n} \frac{n}{n^2 + k} \geq \sum_{k=1}^{n} \frac{1}{n + 1} = n \cdot \frac{1}{n + 1}
\]

Simplifying the right-hand side:

\[
a_n \geq \frac{n}{n + 1}
\]

For large \( n \), \( \frac{n}{n + 1} \to 1 \).

Thus, we have a lower bound:

\[
a_n \geq 1 - \frac{1}{n + 1}
\]

### Step 4: Apply the Sandwich (Squeeze) Theorem
Now, we have the inequalities:

\[
1 - \frac{1}{n + 1} \leq a_n \leq \frac{n^2}{n^2 + 1}
\]

As \( n \to \infty \):

\[
1 - \frac{1}{n + 1} \to 1 \quad \text{and} \quad \frac{n^2}{n^2 + 1} \to 1
\]

By the **sandwich theorem**, we conclude that:

\[
a_n \to 1 \quad \text{as} \quad n \to \infty
\]

Thus, the sequence \( a_n \) is convergent, and its limit is \( 1 \).

---

Would you like more details or have any questions? Here are 5 related questions to expand your understanding:

1. How does the behavior of the general term \( \frac{n}{n^2 + k} \) affect the convergence?
2. Can the sandwich theorem be applied to sequences involving sums in more complex cases?
3. How do we find the upper and lower bounds in different convergence problems?
4. Why is it important that both bounds converge to the same limit in the sandwich theorem?
5. How would this proof change if the terms were \( \frac{n}{n^2 + k^2} \)?

**Tip:** When using the sandwich theorem, the key is finding bounds that closely approximate the behavior of the sequence for large values of \( n \).

Prove an = n/(n^2+1) + n/(n^2+2) .... n/(n^2+n) is a convergent sequence using the sandwich theorem

Explore how to prove the convergence of the sequence a_n = n/(n^2 + 1) + ... + n/(n^2 + n) using the Sandwich Theorem. The solution covers detailed steps in bounding the sequence and applying the theorem for large n, making it suitable for university-level calculus students.

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