Define an → a as n → ∞. Use this to show that (−1)n doesn’t converge to any real
number

To solve this, let's first explore what the notation means and how to apply it to prove that \( (-1)^n \) does not converge to any real number.

### Definition of Convergence

In general, a sequence \( a_n \) converges to a real number \( a \) if:

\[
\lim_{n \to \infty} a_n = a
\]

This means that for every positive number \( \epsilon > 0 \), there exists an integer \( N \) such that for all \( n > N \), the absolute difference between \( a_n \) and \( a \) is less than \( \epsilon \):

\[
|a_n - a| < \epsilon
\]

In simpler terms, as \( n \to \infty \), the terms of the sequence \( a_n \) get arbitrarily close to the value \( a \).

### The Sequence \( (-1)^n \)

Now consider the sequence defined by \( (-1)^n \), which alternates between the values \( 1 \) and \( -1 \):

\[
(-1)^1 = -1, \quad (-1)^2 = 1, \quad (-1)^3 = -1, \quad (-1)^4 = 1, \quad \ldots
\]

This sequence does not "settle down" to a single value as \( n \to \infty \), because it keeps switching between 1 and -1.

### Showing Non-Convergence

Let's suppose, for the sake of contradiction, that \( (-1)^n \) converges to some real number \( a \). According to the definition of convergence, as \( n \to \infty \), the terms of the sequence should get closer and closer to this value \( a \).

1. Consider the subsequence where \( n \) is even:
   \[
   (-1)^{2k} = 1, \quad k = 1, 2, 3, \dots
   \]
   This subsequence converges to \( 1 \).

2. Now consider the subsequence where \( n \) is odd:
   \[
   (-1)^{2k+1} = -1, \quad k = 1, 2, 3, \dots
   \]
   This subsequence converges to \( -1 \).

For the sequence \( (-1)^n \) to converge to some real number \( a \), both subsequences must converge to the same value. However, we see that the even subsequence converges to 1, while the odd subsequence converges to -1. This means that the sequence \( (-1)^n \) does not get arbitrarily close to a single real number.

Thus, the sequence does **not** converge to any real number \( a \).

### Conclusion

The sequence \( (-1)^n \) does not converge to any real number because it alternates between 1 and -1, and the limits of the subsequences (even and odd terms) do not match.

Would you like further details or have any questions?

### Related Questions:
1. What is the formal definition of a convergent sequence?
2. Can you show an example of a sequence that does converge to a real number?
3. What is the difference between a divergent sequence and an oscillating sequence?
4. How does the concept of subsequences help in understanding convergence?
5. What is the importance of limits in mathematical analysis?

### Tip:
When testing for convergence of a sequence, always consider the behavior of subsequences. If two subsequences of the same sequence converge to different values, the sequence itself does not converge.

Define an → a as n → ∞. Use this to show that (−1)^n doesn’t converge to any real number.

Explore the non-convergence of the sequence (-1)^n using the definition of limits and real analysis. Understand why the sequence oscillates and does not approach a single real number. Suitable for undergraduate students studying calculus or real analysis.

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