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How does the inverse relationship between ε and N manifest in the sequence 3n²/(n²−3)?

As the tolerance ε\varepsilon decreases, the required cutoff NN generally increases because the sequence terms must get closer to the limit 3. For ε=0.5\varepsilon = 0.5, N=5N=5. When ε\varepsilon is reduced to 0.150.15, the tolerance band narrows, and points between n=5n=5 and n=7n=7 violate the stricter bound, requiring N=8N=8. However, due to integer rounding, NN does not strictly increase for every infinitesimal decrease in ε\varepsilon; it can stay unchanged over an interval.

Conditions

  • Sequence is an=3n2n2−3a_n = \frac{3n^2}{n^2-3}.
  • Limit is A=3A=3.
  • ε>0\varepsilon > 0.

Reasoning, step by step

  1. Calculate NN for a wider tolerance, e.g., ε=0.5\varepsilon = 0.5, yielding N=5N=5.
  2. Reduce ε\varepsilon to a narrower tolerance, e.g., ε=0.15\varepsilon = 0.15.
  3. Recalculate the threshold: N=⌈9/0.15+3⌉=8N = \lceil \sqrt{9/0.15 + 3} \rceil = 8.
  4. Observe that terms n=5,6,7n=5, 6, 7 fall outside the new band.
  5. Conclude that smaller ε\varepsilon necessitates larger NN, but integer cutoffs may plateau.

Example

For ε=0.5\varepsilon = 0.5, N=5N=5. For ε=0.15\varepsilon = 0.15, N=8N=8. The animation compares wide and narrow tolerance bands.

Common misconceptions

  • Believing that NN must increase strictly every time ε\varepsilon decreases.
  • Thinking that the same NN works for all ε\varepsilon.

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