How does the triangle inequality prove that the sequence $a_n = (-1)^n$ has no limit for any real $L$ ?

Q: How does the triangle inequality prove that the sequence $a_n = (-1)^n$ has no limit for any real $L$?

The triangle inequality shows that for any real number $L$, the sum of the distances from $L$ to $1$ and from $L$ to $-1$ is at least $2$. This implies that at least one of these distances must be greater than or equal to $1$. By choosing $\varepsilon = 1$, we ensure that either the even terms or the odd terms will always be outside the $\varepsilon$-neighborhood of $L$, regardless of $N$. Thus, the sequence cannot converge to any $L$.

The triangle inequality shows that for any real number

L

, the sum of the distances from

L

1

and from

L

-1

is at least

2

. This implies that at least one of these distances must be greater than or equal to

1

. By choosing

\varepsilon = 1

, we ensure that either the even terms or the odd terms will always be outside the

\varepsilon

-neighborhood of

L

, regardless of

N

. Thus, the sequence cannot converge to any

L

Conditions

The sequence is $a_n = (-1)^n$ .
$L$ is an arbitrary real number.
$\varepsilon$ is chosen to be $1$ .

Reasoning, step by step

Assume the sequence converges to some limit $L$ .
Apply the triangle inequality to the distance between the two cluster points: $|1 - (-1)| \le |1 - L| + |L - (-1)|$ .
Simplify the left side: $2 \le |1 - L| + |-1 - L|$ .
Deduce that at least one of the terms $|1 - L|$ or $|-1 - L|$ must be $\ge 1$ .
Choose $\varepsilon = 1$ . If $|1 - L| \ge 1$ , then all even terms $a_{2k}=1$ satisfy $|a_{2k} - L| \ge \varepsilon$ .
If $|-1 - L| \ge 1$ , then all odd terms $a_{2k+1}=-1$ satisfy $|a_{2k+1} - L| \ge \varepsilon$ .
Conclude that for any $N$ , there are terms after $N$ violating the condition, so the sequence diverges.

Example

The script states: 'To rule out every candidate L, the triangle inequality gives $2 \le |1-L| + |-1-L|$ , so at least one class of terms stays at distance at least 1. Such terms occur after every N. Thus $\varepsilon=1$ violates convergence for every real L.'

Common misconceptions

Thinking that checking only $L=1$ and $L=-1$ is sufficient to prove non-convergence.
Misapplying the triangle inequality by assuming equality holds for all $L$ .
Believing that a sequence with two subsequential limits must converge to their average.

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How does the triangle inequality prove that the sequence $a_n = (-1)^n$ has no limit for any real $L$ ?

Conditions

Reasoning, step by step

Example

Common misconceptions

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Connected concepts

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Related questions

How does the triangle inequality prove that the sequence an=(−1)na_n = (-1)^nan​=(−1)n has no limit for any real LLL?

Conditions

Reasoning, step by step

Example

Common misconceptions

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Connected concepts

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How does the triangle inequality prove that the sequence $a_n = (-1)^n$ has no limit for any real $L$ ?