The ε–N Definition

Watch the moment the whole tail of a sequence fits inside the ε\varepsilon-band. That NN is exactly the NN in ∀ε>0 ∃N\forall\varepsilon>0\ \exists N.

∀ε>0 ∃N s.t. n>N⇒∣an−L∣<ε\forall\varepsilon>0\ \exists N\ \text{s.t.}\ n>N\Rightarrow|a_n-L|<\varepsilon

Examples & Exercises

Show that an=1n2a_n = \frac{1}{n^2} converges to 00 from the ε\varepsilon-NN definition, and find NN for ε=0.01\varepsilon = 0.01.

N(ε)=⌈1ε⌉N(\varepsilon) = \left\lceil \frac{1}{\sqrt{\varepsilon}} \right\rceil
an=1/n2a_n = 1/n^2, ε=0.01\varepsilon = 0.01 — the tail enters the band from N=10N = 10.
Show proof
  1. Since ∣1n2−0∣=1n2\left|\frac{1}{n^2} - 0\right| = \frac{1}{n^2}, the inequality 1n2<ε\frac{1}{n^2} < \varepsilon is equivalent to n2>1εn^2 > \frac{1}{\varepsilon}, i.e. n>1εn > \frac{1}{\sqrt{\varepsilon}}.
  2. Put N=⌈1ε⌉N = \left\lceil \frac{1}{\sqrt{\varepsilon}} \right\rceil. If n>Nn > N then n>1εn > \frac{1}{\sqrt{\varepsilon}}, so ∣an−0∣<ε\left|a_n - 0\right| < \varepsilon. Hence an→0a_n \to 0 by the ε\varepsilon-NN definition.
  3. For ε=0.01\varepsilon = 0.01, 10.01=10\frac{1}{\sqrt{0.01}} = 10, so N=10N = 10. Indeed, for n≥11n \geq 11 we have 1n2≤1121<0.01\frac{1}{n^2} \leq \frac{1}{121} < 0.01.

Show that an=nn+1a_n = \frac{n}{n+1} converges to 11, and find NN for ε=0.05\varepsilon = 0.05.

∣nn+1−1∣=1n+1\left|\frac{n}{n+1} - 1\right| = \frac{1}{n+1}
an=n/(n+1)a_n = n/(n+1), ε=0.05\varepsilon = 0.05 — the tail enters the band from N=19N = 19.
Show proof
  1. ∣nn+1−1∣=∣−1n+1∣=1n+1\left|\frac{n}{n+1} - 1\right| = \left|\frac{-1}{n+1}\right| = \frac{1}{n+1}.
  2. 1n+1<ε\frac{1}{n+1} < \varepsilon is equivalent to n+1>1εn + 1 > \frac{1}{\varepsilon}, i.e. n>1ε−1n > \frac{1}{\varepsilon} - 1. Taking NN to be any integer larger than 1ε−1\frac{1}{\varepsilon} - 1 gives ∣an−1∣<ε\left|a_n - 1\right| < \varepsilon for n>Nn > N, so an→1a_n \to 1.
  3. For ε=0.05\varepsilon = 0.05, 10.05−1=19\frac{1}{0.05} - 1 = 19, so N=19N = 19 works: for n≥20n \geq 20, 1n+1≤121<0.05\frac{1}{n+1} \leq \frac{1}{21} < 0.05.

Use ∣(−1)nn∣=1n\left|\frac{(-1)^n}{n}\right| = \frac{1}{n} to prove an=(−1)nn→0a_n = \frac{(-1)^n}{n} \to 0.

∣(−1)nn−0∣=1n<ε  ⟺  n>1ε\left|\frac{(-1)^n}{n} - 0\right| = \frac{1}{n} < \varepsilon \iff n > \frac{1}{\varepsilon}
an=(−1)n/na_n = (-1)^n/n, ε=0.05\varepsilon = 0.05 — despite the sign flips, the tail enters the band from N=20N = 20.
Show proof
  1. ∣(−1)nn−0∣=∣(−1)n∣n=1n\left|\frac{(-1)^n}{n} - 0\right| = \frac{|(-1)^n|}{n} = \frac{1}{n}. The sign (−1)n(-1)^n disappears inside the absolute value.
  2. Given ε>0\varepsilon > 0, take an integer N>1εN > \frac{1}{\varepsilon}. Then n>Nn > N implies 1n<ε\frac{1}{n} < \varepsilon, so ∣an−0∣<ε\left|a_n - 0\right| < \varepsilon.
  3. For ε=0.05\varepsilon = 0.05, 10.05=20\frac{1}{0.05} = 20, so N=20N = 20: for n≥21n \geq 21, 1n≤121<0.05\frac{1}{n} \leq \frac{1}{21} < 0.05.

2. Sequences