How does testing larger indices ($N=20$ vs $N=5$) demonstrate that $a_n=(-1)^n$ does not converge to $L=1$?

Conditions

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

Reasoning, step by step

1. Define the epsilon neighborhood for $L=1$ with $\varepsilon=0.5$: interval $(0.5, 1.5)$.
2. Test $N=5$: Check terms for $n>5$. Odd terms (e.g., $n=7, 9$) are $-1$, falling outside the band.
3. Increase $N$ to 20 to rule out early fluctuations.
4. Check terms for $n>20$. Terms like $n=21, 23$ are still $-1$.
5. Observe that the violation persists indefinitely, proving the 'eventually' condition fails.

Example

At $N=20$, the script highlights orange circles for $n=21, 23$. These points remain far below the target band near 1, showing that moving further right on the graph does not fix the proximity issue.

Common misconceptions

• Thinking that if terms fail initially, they might succeed later.
• Ignoring the alternating nature of the sequence when choosing N.

Watch the explanation

Explore next

• What happens if we choose a smaller epsilon?
• Does this method work for convergent sequences?