to find with the help of subsequences the accumulation points of the sequence an=((-1)^n)( 1+1/n) and to show that they are the only ones.

To find the accumulation points of the sequence \( a_n = (-1)^n \left( 1 + \frac{1}{n} \right) \) using subsequences and show that they are the only ones, we need to analyze the behavior of the sequence by considering its subsequences.

### Step-by-Step Solution:

1. **Define the Sequence**:
   The given sequence is \( a_n = (-1)^n \left( 1 + \frac{1}{n} \right) \).

2. **Identify Subsequences**:
   To find accumulation points, consider subsequences that might converge.

   - **Even-indexed subsequence**:
     Let \( n = 2k \) where \( k \in \mathbb{N} \). Then:
     \[
     a_{2k} = (-1)^{2k} \left( 1 + \frac{1}{2k} \right) = 1 + \frac{1}{2k}
     \]
     As \( k \to \infty \), \( \frac{1}{2k} \to 0 \). Thus, the even-indexed subsequence converges to:
     \[
     \lim_{k \to \infty} a_{2k} = 1
     \]

   - **Odd-indexed subsequence**:
     Let \( n = 2k + 1 \) where \( k \in \mathbb{N} \). Then:
     \[
     a_{2k+1} = (-1)^{2k+1} \left( 1 + \frac{1}{2k+1} \right) = -\left( 1 + \frac{1}{2k+1} \right)
     \]
     As \( k \to \infty \), \( \frac{1}{2k+1} \to 0 \). Thus, the odd-indexed subsequence converges to:
     \[
     \lim_{k \to \infty} a_{2k+1} = -1
     \]

3. **Determine Accumulation Points**:
   An accumulation point of a sequence is a point such that every neighborhood of the point contains infinitely many terms of the sequence. From the subsequences above, we see:
   \[
   \lim_{k \to \infty} a_{2k} = 1 \quad \text{and} \quad \lim_{k \to \infty} a_{2k+1} = -1
   \]
   Thus, the accumulation points of the sequence \( \{a_n\} \) are \( 1 \) and \( -1 \).

4. **Show These Are the Only Accumulation Points**:
   We need to prove that no other values can be accumulation points. Suppose there exists some other accumulation point \( L \). By definition, for \( L \) to be an accumulation point, there must be a subsequence \( a_{n_k} \) that converges to \( L \).

   - Given the structure of \( a_n = (-1)^n \left( 1 + \frac{1}{n} \right) \), the terms alternate between positive and negative values.
   - The even-indexed subsequence converges to 1, and the odd-indexed subsequence converges to -1.
   - Since \( a_n \) only takes values close to 1 or -1 as \( n \to \infty \), the only possible limits of any subsequence must be either 1 or -1.

   Therefore, no other accumulation points exist.

### Conclusion:
The accumulation points of the sequence \( a_n = (-1)^n \left( 1 + \frac{1}{n} \right) \) are \( 1 \) and \( -1 \). These are the only accumulation points because the subsequences converge to these points exclusively.

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Would you like more details or have any questions? Here are five related questions you might consider asking:

1. How do we formally prove the convergence of a subsequence?
2. What are accumulation points and why are they important in sequence analysis?
3. Can you explain how to find accumulation points for other types of sequences?
4. What are the different types of limits a sequence can have?
5. How do accumulation points relate to the concept of the limit superior and limit inferior?

**Tip:** When analyzing sequences, always consider breaking them into subsequences to understand their behavior better. This can reveal hidden patterns and accumulation points.

Explore how to find accumulation points of the sequence \( a_n = (-1)^n \left( 1 + \frac{1}{n} \right) \) using subsequences and prove they are the only ones. Understand the behavior of alternating signs and convergence patterns. Suitable for advanced undergraduate mathematics.

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