Lim sup and Lim inf

Where the suprema and infima of tails settle: for an=(−1)n(1+1/n)a_n = (-1)^n(1+1/n), lim sup⁡an=1\limsup a_n = 1 and lim inf⁡an=−1\liminf a_n = -1. lim sup⁡an=lim inf⁡an  ⟺  lim⁡an\limsup a_n = \liminf a_n \iff \lim a_n exists.

lim sup⁡an=lim⁡N→∞sup⁡n≥Nan\limsup a_n = \lim_{N\to\infty}\sup_{n\ge N} a_n
lim inf⁡an=lim⁡N→∞inf⁡n≥Nan\liminf a_n = \lim_{N\to\infty}\inf_{n\ge N} a_n
lim sup⁡an=lim inf⁡an  ⟺  lim⁡an exists\limsup a_n = \liminf a_n \iff \lim a_n \text{ exists}

Examples & Exercises

Find lim sup⁡an\limsup a_n and lim inf⁡an\liminf a_n of an=(−1)n(1+1n)a_n = (-1)^n\left(1+\frac{1}{n}\right).

a2k=1+12k→1,a2k+1=−(1+12k+1)→−1a_{2k} = 1+\frac{1}{2k} \to 1, \qquad a_{2k+1} = -\left(1+\frac{1}{2k+1}\right) \to -1
n=24n=24: even term a24=2524≈1.0417a_{24} = \frac{25}{24} \approx 1.0417, odd term a23=−2423≈−1.0435a_{23} = -\frac{24}{23} \approx -1.0435.
Show proof
  1. The even subsequence a2k=1+12ka_{2k} = 1+\frac{1}{2k} decreases to 11 (starting at a2=1.5a_2 = 1.5).
  2. The odd subsequence a2k+1=−(1+12k+1)a_{2k+1} = -\left(1+\frac{1}{2k+1}\right) increases to −1-1 (starting at a3=−43a_3 = -\frac{4}{3}).
  3. Compute the tail suprema directly. sup⁡n≥Nan=max⁡(sup⁡n≥Nn evenan,  sup⁡n≥Nn oddan)\sup_{n \ge N} a_n = \max\left(\sup_{\substack{n \ge N \\ n \text{ even}}} a_n,\; \sup_{\substack{n \ge N \\ n \text{ odd}}} a_n\right). On the even tail, a2k=1+12k↓1a_{2k} = 1 + \frac{1}{2k} \downarrow 1, so sup⁡even ≥Nan=1+12⌈N/2⌉\sup_{\text{even } \ge N} a_n = 1 + \frac{1}{2\lceil N/2 \rceil}. On the odd tail every term is at most −1-1, so sup⁡odd ≥Nan≤−1<1\sup_{\text{odd } \ge N} a_n \le -1 < 1. Hence the max⁡\max is decided by the even side: sup⁡n≥Nan=1+12⌈N/2⌉\sup_{n \ge N} a_n = 1 + \frac{1}{2\lceil N/2 \rceil}. Given ε>0\varepsilon > 0, N>1εN > \frac{1}{\varepsilon} gives 0≤sup⁡n≥Nan−1=12⌈N/2⌉≤1N<ε0 \le \sup_{n \ge N} a_n - 1 = \frac{1}{2\lceil N/2 \rceil} \le \frac{1}{N} < \varepsilon — so sup⁡n≥Nan→1\sup_{n \ge N} a_n \to 1 and lim sup⁡an=1\limsup a_n = 1.
  4. The infimum is symmetric. In inf⁡n≥Nan=min⁡(inf⁡even ≥Nan,  inf⁡odd ≥Nan)\inf_{n \ge N} a_n = \min\left(\inf_{\text{even } \ge N} a_n,\; \inf_{\text{odd } \ge N} a_n\right) the even side has inf⁡≥1>−1\inf \ge 1 > -1, while on the odd side a2k+1=−(1+12k+1)↑−1a_{2k+1} = -\left(1 + \frac{1}{2k+1}\right) \uparrow -1, so inf⁡odd ≥Nan=−(1+12⌊N/2⌋+1)\inf_{\text{odd } \ge N} a_n = -\left(1 + \frac{1}{2\lfloor N/2 \rfloor + 1}\right). Hence inf⁡n≥Nan=−(1+12⌊N/2⌋+1)→−1\inf_{n \ge N} a_n = -\left(1 + \frac{1}{2\lfloor N/2 \rfloor + 1}\right) \to -1 and lim inf⁡an=−1\liminf a_n = -1.

Prove that if lim⁡an=L\lim a_n = L exists, then lim sup⁡an=lim inf⁡an=L\limsup a_n = \liminf a_n = L.

L−ε<an<L+ε(n≥N)  ⟹  L−ε≤lim inf⁡an≤lim sup⁡an≤L+εL-\varepsilon < a_n < L+\varepsilon \quad (n \ge N) \implies L-\varepsilon \le \liminf a_n \le \limsup a_n \le L+\varepsilon
n=24n=24: this scene's sequence does not converge, so lim sup⁡=1\limsup = 1 differs from lim inf⁡=−1\liminf = -1.
Show proof
  1. Let ε>0\varepsilon > 0. Since lim⁡an=L\lim a_n = L, there is NN with ∣an−L∣<ε|a_n - L| < \varepsilon, i.e. L−ε<an<L+εL - \varepsilon < a_n < L + \varepsilon, for all n≥Nn \ge N.
  2. Taking sup and inf over the tail n≥Nn \ge N gives L−ε≤inf⁡n≥Nan≤sup⁡n≥Nan≤L+εL - \varepsilon \le \inf_{n\ge N} a_n \le \sup_{n\ge N} a_n \le L + \varepsilon. Letting N→∞N \to \infty, L−ε≤lim inf⁡an≤lim sup⁡an≤L+εL - \varepsilon \le \liminf a_n \le \limsup a_n \le L + \varepsilon.
  3. Since ε>0\varepsilon > 0 is arbitrary, lim inf⁡an=lim sup⁡an=L\liminf a_n = \limsup a_n = L.

Prove the root test: if ρ=lim sup⁡∣an∣1/n<1\rho = \limsup |a_n|^{1/n} < 1, then ∑an\sum a_n converges absolutely.

ρ=lim sup⁡∣an∣1/n<1  ⟹  ∣an∣≤rn eventually  ⟹  ∑∣an∣ converges\rho = \limsup |a_n|^{1/n} < 1 \implies |a_n| \le r^n \text{ eventually} \implies \sum |a_n| \text{ converges}
n=24n=24: ∣a24∣1/24≈1.0017|a_{24}|^{1/24} \approx 1.0017 and lim sup⁡∣an∣1/n=1\limsup |a_n|^{1/n} = 1 — the inconclusive boundary.
Show proof
  1. Pick rr with ρ<r<1\rho < r < 1. Since ρ\rho is the limit of the tail suprema, there is NN such that ∣an∣1/n≤r|a_n|^{1/n} \le r, i.e. ∣an∣≤rn|a_n| \le r^n, for all n≥Nn \ge N.
  2. Since 0≤r<10 \le r < 1, the geometric series ∑rn\sum r^n converges. By comparison, ∑∣an∣\sum |a_n| converges, so ∑an\sum a_n converges absolutely.

3. Sequences