Stolz–Cesàro Theorem | Real Analysis

When the full ratio is stubborn, this scene detours through the ratio of differences. It is the sequence version of l'Hôpital's rule.

Compare the growth of single steps, and that ratio settles first. The full ratio follows slowly from behind. The denominator must increase strictly and grow without bound. Both ratios share the same destination. The headline example's limit is the second formula below.

For example, on a 🚗 car trip the speed over the last minute settles at 60 km/h. Then the average speed over the whole trip gets dragged toward 60 km/h as well.

Examples & Exercises

If $a_n \to L$ , prove with the Stolz–Cesàro theorem that the Cesàro mean $\frac{a_1+\cdots+a_n}{n}$ also converges to $L$ (take $b_n = n$ ).

\lim_{n\to\infty}\frac{a_1+\cdots+a_n}{n} = L

n=100

: in this scene's example the difference ratio

r_{100} \approx 0.3367

settles at

\frac{1}{3}

first and the full ratio

s_{100} \approx 0.3384

follows. The Cesàro mean is the special case

b_n = n

, where the difference ratio is

a_{n+1}

itself.

Show proof

Let $c_n = a_1+\cdots+a_n$ and $b_n = n$ . The sequence $(b_n)$ is strictly increasing and unbounded, so it satisfies the hypotheses of the Stolz–Cesàro theorem.
The difference ratio is $\frac{c_{n+1}-c_n}{b_{n+1}-b_n} = \frac{a_{n+1}}{1} = a_{n+1}$ . A shift of a convergent sequence converges to the same limit: if $a_n \to L$ , then an $N$ for a given $\varepsilon > 0$ also works for $a_{n+1}$ (since $n \ge N \Rightarrow n+1 \ge N$ , giving $|a_{n+1} - L| < \varepsilon$ ).
Hence $a_{n+1} \to L$ , and the Stolz–Cesàro theorem gives $\frac{c_n}{b_n} = \frac{a_1+\cdots+a_n}{n} \to L$ .

Use the Stolz–Cesàro theorem to prove $\displaystyle \lim_{n\to\infty}\frac{1^2+2^2+\cdots+n^2}{n^3} = \frac{1}{3}$ .

\lim_{n\to\infty}\frac{1^2+2^2+\cdots+n^2}{n^3} = \frac{1}{3}

n=200

r_{200} \approx 0.3350

s_{200} \approx 0.3358

— both ratios have nearly reached

\frac{1}{3}

Show proof

Let $a_n = 1^2+\cdots+n^2$ and $b_n = n^3$ . The sequence $(b_n)$ is strictly increasing for $n \ge 1$ and unbounded.
Since $a_{n+1}-a_n = (n+1)^2$ and $b_{n+1}-b_n = (n+1)^3-n^3 = 3n^2+3n+1$ , the difference ratio is $\frac{(n+1)^2}{3n^2+3n+1} = \frac{1+2/n+1/n^2}{3+3/n+1/n^2} \to \frac{1}{3}$ .
By the Stolz–Cesàro theorem, $\frac{a_n}{b_n} = \frac{1^2+\cdots+n^2}{n^3} \to \frac{1}{3}$ .

$1+3+5+\cdots+(2n-1)$ is the sum of the first $n$ odd numbers.

Use the Stolz–Cesàro theorem to prove $\displaystyle \lim_{n\to\infty}\frac{1+3+\cdots+(2n-1)}{n^2} = 1$ .

\sum_{k=1}^{n}(2k-1) = n^2, \qquad \lim_{n\to\infty}\frac{1+3+\cdots+(2n-1)}{n^2} = 1

n=60

: in this scene's example

r_{60} \approx 0.3389

s_{60} \approx 0.3417

— the odd-sum problem runs on the same machinery (

b_n = n^2

, difference ratio exactly

1

Show proof

Let $a_n = 1+3+\cdots+(2n-1)$ ; by induction $a_n = n^2$ : for $n=1$ , $1 = 1^2$ , and if $a_n = n^2$ then $a_{n+1} = n^2+(2n+1) = (n+1)^2$ . The sequence $b_n = n^2$ is strictly increasing and unbounded.
Since $a_{n+1}-a_n = 2n+1$ and $b_{n+1}-b_n = (n+1)^2-n^2 = 2n+1$ , the difference ratio is $\frac{2n+1}{2n+1} = 1$ for every $n$ , hence its limit is $1$ .
By the Stolz–Cesàro theorem, $\frac{a_n}{b_n} \to 1$ — in fact $a_n = b_n$ , so the ratio is exactly $1$ for every $n$ .