Lim sup and Lim inf

Watch only the tails of a sequence and track their highest and lowest values. The scene checks where the two of them settle.

The supremum of the tails keeps stepping down, and the infimum keeps stepping up. The upper settling point is the lim sup, the lower one the lim inf. If the two meet, the limit exists. An oscillating sequence keeps them apart forever. The precise definitions are the formulas below.

For example, record each day's high and low temperature with a 🌡️ thermometer. As days pass, the highs gather near one value and the lows near another. If the two values coincide, the weather has settled on a single temperature.

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. Boundedness first: a convergent sequence is bounded — with ε=1\varepsilon = 1 there is N0N_0 with ∣an−L∣<1|a_n - L| < 1 for n≥N0n \ge N_0, so ∣an∣≤∣L∣+1|a_n| \le |L| + 1 on the tail, and the finitely many earlier terms are bounded by their maximum; hence ∣an∣≤M|a_n| \le M for all nn and some MM.

    Therefore every tail supremum sup⁡n≥Nan\sup_{n\ge N} a_n and tail infimum inf⁡n≥Nan\inf_{n\ge N} a_n is finite, and lim sup⁡an,lim inf⁡an\limsup a_n, \liminf a_n are well-defined real numbers.

    Now 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 ∣an∣1/n≥0|a_n|^{1/n} \ge 0 for every nn, ρ=lim sup⁡∣an∣1/n≥0\rho = \limsup |a_n|^{1/n} \ge 0; hence the chosen rr satisfies 0≤ρ<r<10 \le \rho < r < 1, in particular 0≤r<10 \le r < 1, so the geometric series ∑rn\sum r^n converges. By comparison, ∑∣an∣\sum |a_n| converges, so ∑an\sum a_n converges absolutely.

3. Sequences