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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.

Determine whether the sequence (-1)^{n+1} * (n / (2n+1)) from n=1 to n=∞ converges or diverges.

Explore whether the sequence (-1)^(n+1) * (n / (2n+1)) converges or diverges. The solution involves limits, alternating sequences, and the divergence test. Suitable for students studying Calculus I or II.

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