Disproving a sequence limit from the definition

The proof uses the negation of the formal definition of convergence. It establishes that for every real number $L$, there exists an $\varepsilon > 0$ (specifically $\varepsilon=1$) such that for every natural number $N$, there exists an index $n > N$ where $|a_n - L| \ge \varepsilon$.

Conditions: The sequence is $a_n = (-1)^n$.; The definition of convergence is $\forall \varepsilon > 0, \exists N, \forall n > N: |a_n - L| < \varepsilon$.; The negation is $\forall L, \exists \varepsilon > 0, \forall N, \exists n > N: |a_n - L| \ge \varepsilon$.

The divergence is proven by asserting: For every real number $L$, there exists an $\varepsilon > 0$ such that for every integer $N$, there is an index $n > N$ where $|a_n - L| \ge \varepsilon$.

Conditions: Standard definition of sequence convergence is assumed.; $a_n = (-1)^n$.

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.

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

The sequence fails to converge to $L=-1$ because the even-indexed terms remain at $1$, which is a fixed distance away from $-1$. Similar to the case for $L=1$, no matter how large the cutoff index $N$ is, there will always be subsequent even terms that fall outside the tolerance band centered at $-1$, violating the convergence condition.

Conditions: The sequence is defined as $a_n = (-1)^n$.; The candidate limit is $L=-1$.; The tolerance $\varepsilon$ is chosen such that $0 < \varepsilon < 2$ (e.g., $\varepsilon=0.5$).

A fundamental theorem states that if a sequence converges to $L$, every subsequence must also converge to the same $L$. Here, odd terms approach $-1$ and even terms approach $1$.

Conditions: Theorem: Convergence implies unique subsequential limits.; Odd subsequence limit is $-1$.; Even subsequence limit is $1$.

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 $1$.

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

Increasing $N$ tests whether the sequence eventually settles into the tolerance band around $L=1$. For $a_n=(-1)^n$, odd-indexed terms remain at $-1$, which is outside the band $(0.5, 1.5)$ regardless of how large $N$ becomes.

Conditions: Candidate limit $L=1$.; Tolerance $\varepsilon=0.5$ (band width 1.0).; Sequence term $a_n = (-1)^n$.

Rejecting specific values like 1 or -1 only proves those particular numbers are not limits. To rigorously prove the limit does not exist, one must show that *every* possible real number $L$ fails the convergence definition.

Conditions: The sequence is defined as $a_n = (-1)^n$.; Proof requires ruling out all $L \in \mathbb{R}$.

The sequence fails to converge to $L=1$ because for any chosen tolerance $\varepsilon$, the odd-indexed terms remain at $-1$, which is a fixed distance away from $1$. No matter how large the cutoff index $N$ is, there will always be subsequent odd terms that fall outside the tolerance band centered at $1$, violating the requirement that all terms after $N$ must stay within $\varepsilon$ of the limit.

Conditions: The sequence is defined as $a_n = (-1)^n$.; The candidate limit is $L=1$.; The tolerance $\varepsilon$ is chosen such that $0 < \varepsilon < 2$ (e.g., $\varepsilon=0.5$).