How does testing the alternative limit $L=-1$ reveal failure for the even-indexed terms of $a_n=(-1)^n$ ?

Q: How does testing the alternative limit $L=-1$ reveal failure for the even-indexed terms of $a_n=(-1)^n$?

When assuming $L=-1$ with $\varepsilon=0.5$, the valid range is $(-1.5, -0.5)$. Even-indexed terms equal $1$, which lies well above this upper bound. Like the case for $L=1$, these violations persist for arbitrarily large $N$.

When assuming

L=-1

with

\varepsilon=0.5

, the valid range is

(-1.5, -0.5)

. Even-indexed terms equal

1

, which lies well above this upper bound. Like the case for

L=1

, these violations persist for arbitrarily large

N

Conditions

Candidate limit $L=-1$ .
Tolerance $\varepsilon=0.5$ .
Sequence $a_n = (-1)^n$ .

Reasoning, step by step

Set center line at $L=-1$ with green band width $2\varepsilon=1.0$ .
Identify even indices ( $n=2, 4, \dots$ ) where $a_n = 1$ .
Check position relative to band $(-1.5, -0.5)$ .
Observe that $1 > -0.5$ , so even terms are outside.
Verify for large $N$ (e.g., $N=25$ ): terms like $n=26, 28$ remain at $1$ , failing to enter the neighborhood.

Example

The script notes: 'Trying larger $N=25$ , terms like $n=26$ , 28 still hover high up, refusing to enter the neighborhood of -1.' This mirrors the failure seen at $L=1$ but for the opposite subsequence.

Common misconceptions

Assuming that because odd terms fit $L=-1$ , the whole sequence converges.
Overlooking the contribution of even terms to the overall behavior.

Watch the explanation

BilibiliDisproving a sequence limit from the definition

0:44 – 1:18Watch this moment ↗

Explore next

Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.

How does testing the alternative limit L=−1L=-1L=−1 reveal failure for the even-indexed terms of an=(−1)na_n=(-1)^nan​=(−1)n?