Stolz-Cesàro Theorem
May 8, 2008 in Elementary Math Problem Solving, Problem Corner, Some theorems | Tags: cesaro, limit, stolz | by Vishal
The following theorem, I feel, is not very well-known, though it is a particularly useful one for solving certain types of “limit” problems. Let me pose a couple of elementary problems and offer their solutions. First, the theorem.
Stolz-Cesàro: Let (a_n)_{n \ge 1} and (b_n)_{n \ge 1} be two sequences of real numbers, such that (b_n) is positive, strictly increasing and unbounded. Then,
\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n},
if the limit on the right hand side exists.
The proof involves the usual \epsilon - \delta method, and I will avoid presenting it here since it isn’t particularly interesting. Just as Abel’s lemma is the discrete analogue of integration by parts, the Stolz-Cesàro theorem may be considered the discrete analogue of L’Hospital’s rule in calculus.
Problem 1: Evaluate the limit \displaystyle \lim_{n \to \infty} \frac{1^k + 2^k + \ldots + n^k}{n^{k+1}}, where k \in \mathbb{N}.
Solution: One may certainly consider the above limit as a Riemann-sum which may then be transformed into the integral \displaystyle \int_0^1 x^k \, dx, which then obviously evaluates to 1/(k+1). But, we will take a different route here.
First, let a_n = 1^k + 2^k + \ldots + n^k and b_n = n^{k+1}. Then, we note that the sequence (b_n) is positive, strictly increasing and unbounded. Now,
\displaystyle \lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = \lim_{n \to \infty} \frac{(n+1)^k}{(n+1)^{k+1} - n^{k+1}}
\displaystyle = \lim_{n \to \infty} \frac{(n+1)^k}{\left(1 + \binom{k+1}{1}n + \binom{k+1}{2}n^2 + \ldots + \binom{k+1}{k}n^k + n^{k+1}\right) - n^{k+1}}
(using the binomial theorem)
\displaystyle = \lim_{n \to \infty} \frac{(n+1)^k / n^k}{\left(1 + \binom{k+1}{1}n + \binom{k+1}{2}n^2 + \ldots + \binom{k+1}{k}n^k \right) / n^k}
\displaystyle = \lim_{n \to \infty} \frac{(1 + 1/n)^k}{\binom{k+1}{k}} = \frac1{k+1}.
Therefore, using the Stolz-Cesàro theorem, we conclude that the required limit is also 1/(k+1).
Let us now look at another problem where applying the aforesaid theorem makes our job a lot easier. This problem is an example of one that is not amenable to the other usual methods of evaluating limits.
Problem 2: Let k\geq 2 be integers and suppose C: y = \sqrt {2x + 1}\ (x > 0). Given the tangent line at the point (a_{k},\, \sqrt {2a_{k} + 1}) from the point (0, k) to C, evaluate
\displaystyle \lim_{n\to\infty}\frac {1}{n^{3}}\sum_{k = 2}^{n}a_{k}.
Solution:(This is basically the solution I had offered elsewhere a while ago; so, it’s pretty much copy/paste!)
\displaystyle y = \sqrt {2x + 1}\Rightarrow \frac {dy}{dx} = \frac {1}{\sqrt {2x + 1}}.
So, the equation of the tangent line at the point (a_{k}, \sqrt {2a_{k} + 1}) is given by
\displaystyle y - \sqrt {2a_{k} + 1} = \frac {1}{\sqrt {2a_{k} + 1}}(x - a_{k}).
Since the point (0,k) lies on this line, we must have
\displaystyle k - \sqrt {2a_{k} + 1} = \frac {1}{\sqrt {2a_{k} + 1}}( - a_{k})
The above, after squaring and some algebraic manipulation yields
a_{k}^{2} + 2(1 - k^{2})a_{k} + 1 - k^{2} = 0, which implies a_{k} = (k^{2} - 1) + k(\sqrt {k^{2} - 1}). We drop the negative root because a_{k} > 0 for all k\ge 2.
(This is where the Stolz-Cesàro theorem actually comes into play!)
Now, let (b_{n}) and (c_{n}) be two sequences such that \displaystyle b_{n} = \sum_{k = 2}^{n}a_{k} and c_{n} = n^{3}.
Note that (c_{n}) is a positive, increasing and unbounded sequence.
Therefore, \displaystyle \lim_{n\to \infty}\frac {b_{n + 1} - b_{n}}{c_{n + 1} - c_{n}} = \lim_{n\rightarrow \infty}\frac {\sum_{k = 2}^{n + 1}a_{k} - \sum_{k = 2}^{n}a_{k}}{(n + 1)^{3} - n^{3}} = \lim_{n\rightarrow \infty}\frac {a_{n + 1}}{3n^{2} + 3n + 1}
\displaystyle = \lim_{n\to \infty}\frac {(n + 1)^{2} - 1 + (n + 1)\sqrt {(n + 1)^{2} - 1}}{3n^{2} + 3n + 1}
\displaystyle = \lim_{n \to \infty} \frac{(1 + 1/n)^2 - 1/n^2 + (1 + 1/n)\sqrt{1 + 2/n}}{3 + 3/n + 1/n^2}
= 2/3.
Therefore, by the Stolz- Cesàro theorem, we have
\displaystyle \lim_{n\to \infty}\frac {b_{n}}{c_{n}} = 2/3 \,, and so
\displaystyle \lim_{n\to \infty}\frac {1}{n^{3}}\sum_{k = 2}^{n}a_{k} = 2/3.
19 comments
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May 8, 2008 at 5:15 pm
John Armstrong
cite
September 19, 2012 at 10:41 am
Luqing Ye
Abel’s lemma can be used to prove integration by parts.Stolz’s theorem can be used to prove L’Hospital’s law.
May 8, 2008 at 11:40 pm
Todd Trimble
Re the proof of Stolz-Cesàro: there’s the fun little fact that if b, d > 0 and \frac{a}{b} < \frac{c}{d}, then
\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}.
(“the quotient sum of two fractions is between the two fractions”). It follows that if the fractions \frac{a_{n+1}-a_n}{b_{n+1}-b_n} are in the interval (L - \epsilon, L + \epsilon) for n \geq N, then so are all the “telescoping quotient sums” \frac{a_{n+m} - a_n}{b_{n+m} - b_n}. Since the b_{n+m} grow without bound, \frac{a_{n+m}}{b_{n+m}} is also in this interval for m sufficiently large. The conclusion follows.
May 9, 2008 at 1:47 am
Vishal Lama
Todd,
That is a wonderful proof! Thanks for sharing it. I hope you won’t mind if I incorporate the proof in the post some time later. I may perhaps expand a few steps just so that some of our readers may find it easier to see what you just did.
May 9, 2008 at 7:40 pm
John Armstrong
Another consequence of the lemma Todd states is that batting averages actually behave like averages, in that they lie in the middle of the data.
May 10, 2008 at 1:14 am
misha
There is still another neat way to figure out the limit in problem 1. It is based on the combinatorial identity \binom{k}{k} + \binom{k+1}{k} + \dots \binom{n-1}{k} =\binom{n}{k+1} that says that to pick k+1 numbers from 1,2,3,…,n you need first to pick your biggest number and then the rest k of them.
May 10, 2008 at 1:43 am
misha
Still another proof of Stolz-Cesàro. Extend the sequences to piecewise-linear functions by linearly interpolatiing between n and n+1. Then you can either notice that L’Hospital’s rule holds for comtinuous piecewise-differentiable functions, or smooth out the corners and use L’Hospital directly.
May 10, 2008 at 5:02 am
Vishal Lama
Misha,
Those are awesome proofs! Thanks very much!
May 10, 2008 at 7:03 am
misha
Still another way to look at Stolz-Cesàro, that makes it sort of obvious. Begin by linear interpolation, like in my previous comment, then take b as the independent variable.
May 11, 2008 at 6:47 am
misha
That made me think about the alternative proof of the L’Hospitals’s rule itself. Just take the denominator as the independent variable etc.
May 16, 2008 at 4:35 am
Ars Mathematica » Blog Archive » Stolz-Cesaro Theorem
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