Stolz–Cesàro Theorem

The sequence version of l'Hôpital: if (bn)(b_n) is strictly increasing and unbounded and lim⁡an+1−anbn+1−bn=L\lim \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = L, then lim⁡anbn=L\lim \frac{a_n}{b_n} = L. Watch the difference ratio settle at 13\frac{1}{3} first and the full ratio follow for an=12+⋯+n2a_n = 1^2+\cdots+n^2, bn=n3b_n = n^3.

lim⁡an+1−anbn+1−bn=L  ⟹  lim⁡anbn=L\lim \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = L \implies \lim \frac{a_n}{b_n} = L
lim⁡n→∞12+⋯+n2n3=13\lim_{n\to\infty}\frac{1^2+\cdots+n^2}{n^3} = \frac{1}{3}

Examples & Exercises

If an→La_n \to L, prove with the Stolz–Cesàro theorem that the Cesàro mean a1+⋯+ann\frac{a_1+\cdots+a_n}{n} also converges to LL (take bn=nb_n = n).

lim⁡n→∞a1+⋯+ann=L\lim_{n\to\infty}\frac{a_1+\cdots+a_n}{n} = L
n=100n=100: in this scene's example the difference ratio r100≈0.3367r_{100} \approx 0.3367 settles at 13\frac{1}{3} first and the full ratio s100≈0.3384s_{100} \approx 0.3384 follows. The Cesàro mean is the special case bn=nb_n = n, where the difference ratio is an+1a_{n+1} itself.
Show proof
  1. Let cn=a1+⋯+anc_n = a_1+\cdots+a_n and bn=nb_n = n. The sequence (bn)(b_n) is strictly increasing and unbounded, so it satisfies the hypotheses of the Stolz–Cesàro theorem.
  2. The difference ratio is cn+1−cnbn+1−bn=an+11=an+1→L\frac{c_{n+1}-c_n}{b_{n+1}-b_n} = \frac{a_{n+1}}{1} = a_{n+1} \to L. Hence the Stolz–Cesàro theorem gives cnbn=a1+⋯+ann→L\frac{c_n}{b_n} = \frac{a_1+\cdots+a_n}{n} \to L.

Use the Stolz–Cesàro theorem to prove lim⁡n→∞12+22+⋯+n2n3=13\displaystyle \lim_{n\to\infty}\frac{1^2+2^2+\cdots+n^2}{n^3} = \frac{1}{3}.

lim⁡n→∞12+22+⋯+n2n3=13\lim_{n\to\infty}\frac{1^2+2^2+\cdots+n^2}{n^3} = \frac{1}{3}
n=200n=200: r200≈0.3350r_{200} \approx 0.3350, s200≈0.3358s_{200} \approx 0.3358 — both ratios have nearly reached 13\frac{1}{3}.
Show proof
  1. Let an=12+⋯+n2a_n = 1^2+\cdots+n^2 and bn=n3b_n = n^3. The sequence (bn)(b_n) is strictly increasing for n≥1n \ge 1 and unbounded.
  2. Since an+1−an=(n+1)2a_{n+1}-a_n = (n+1)^2 and bn+1−bn=(n+1)3−n3=3n2+3n+1b_{n+1}-b_n = (n+1)^3-n^3 = 3n^2+3n+1, the difference ratio is (n+1)23n2+3n+1=1+2/n+1/n23+3/n+1/n2→13\frac{(n+1)^2}{3n^2+3n+1} = \frac{1+2/n+1/n^2}{3+3/n+1/n^2} \to \frac{1}{3}.
  3. By the Stolz–Cesàro theorem, anbn=12+⋯+n2n3→13\frac{a_n}{b_n} = \frac{1^2+\cdots+n^2}{n^3} \to \frac{1}{3}.

1+3+5+⋯+(2n−1)1+3+5+\cdots+(2n-1) is the sum of the first nn odd numbers. Use the Stolz–Cesàro theorem to prove lim⁡n→∞1+3+⋯+(2n−1)n2=1\displaystyle \lim_{n\to\infty}\frac{1+3+\cdots+(2n-1)}{n^2} = 1.

∑k=1n(2k−1)=n2,lim⁡n→∞1+3+⋯+(2n−1)n2=1\sum_{k=1}^{n}(2k-1) = n^2, \qquad \lim_{n\to\infty}\frac{1+3+\cdots+(2n-1)}{n^2} = 1
n=60n=60: in this scene's example r60≈0.3389r_{60} \approx 0.3389, s60≈0.3417s_{60} \approx 0.3417 — the odd-sum problem runs on the same machinery (bn=n2b_n = n^2, difference ratio exactly 11).
Show proof
  1. Let an=1+3+⋯+(2n−1)a_n = 1+3+\cdots+(2n-1); by induction an=n2a_n = n^2: for n=1n=1, 1=121 = 1^2, and if an=n2a_n = n^2 then an+1=n2+(2n+1)=(n+1)2a_{n+1} = n^2+(2n+1) = (n+1)^2. The sequence bn=n2b_n = n^2 is strictly increasing and unbounded.
  2. Since an+1−an=2n+1a_{n+1}-a_n = 2n+1 and bn+1−bn=(n+1)2−n2=2n+1b_{n+1}-b_n = (n+1)^2-n^2 = 2n+1, the difference ratio is 2n+12n+1=1\frac{2n+1}{2n+1} = 1 for every nn, hence its limit is 11.
  3. By the Stolz–Cesàro theorem, anbn→1\frac{a_n}{b_n} \to 1 — in fact an=bna_n = b_n, so the ratio is exactly 11 for every nn.

2. Sequences