The given sequence is:

$a_n = (-1)^{n+1} \frac{n}{2n+1}$

We are asked to determine whether the sequence converges or diverges as $n \to \infty$.

Step 1: Check the Limit of $a_n$ as $n \to \infty$

We start by evaluating the limit of $a_n$ as $n \to \infty$.

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left( (-1)^{n+1} \frac{n}{2n+1} \right)$

First, let's simplify the expression $\frac{n}{2n+1}$.

$\frac{n}{2n+1} = \frac{n}{2n \left(1 + \frac{1}{2n} \right)} \approx \frac{n}{2n} = \frac{1}{2} \quad \text{as} \quad n \to \infty$

Thus,

$\lim_{n \to \infty} \frac{n}{2n+1} = \frac{1}{2}$

Now, we consider the alternating factor $(-1)^{n+1}$. Since this term alternates between 1 and -1 as $n$ increases, the sequence does not settle on a single value.

Step 2: Test for Convergence

For a sequence to converge, the absolute value of the sequence must approach a limit. However, due to the alternating nature of $(-1)^{n+1}$, the sequence alternates between positive and negative values, and:

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} (-1)^{n+1} \cdot \frac{1}{2} = \pm \frac{1}{2}$

This implies that the sequence does not approach a single value but oscillates between $\frac{1}{2}$ and $-\frac{1}{2}$. Therefore, the sequence diverges.

Conclusion:

The sequence $a_n = (-1)^{n+1} \frac{n}{2n+1}$ diverges as $n \to \infty$.

Would you like more details or have any questions?

Here are some related questions to think about:

1. What is the difference between the convergence of a series and a sequence?
2. How do we determine the limit of an alternating sequence?
3. What are other common methods to check the convergence of a sequence?
4. Can a sequence oscillate and still converge under certain conditions?
5. How does the form of the denominator influence the behavior of a sequence as $n \to \infty$?

Tip: Always check the behavior of alternating sequences carefully; they can sometimes seem bounded but may still diverge if they fail to approach a single value.