Stolz Cesaro
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landau911Stolz Cesaro
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landau911c Gabriel Nagy
The Stolz-Cesaro Theorem
The Theorem. If (bn ) is a sequence of positive real numbers, such that n=1 then for any sequence (an ) R one has the inequalities: n=1 lim sup a1 + a2 + + an an lim sup ; n b1 + b2 + + bn n bn an a1 + a2 + + an lim inf . lim inf n bn n b1 + b2 + + bn
n=1 bn
= ,
(1) (2)
In particular, if the sequence (an /bn ) has a limit, then n=1 a1 + a2 + + an an = lim . n bn n b1 + b2 + + bn lim
Proof. . It is quite clear that we only need to prove (1), since the other inequality follows by replacing an with an . The inequality (1) is trivial, if the right-hand side is +. Assume then that the quantity L = lim supn (an /bn ) is either nite or , and let us x for the moment some number > L. By the denition of lim sup, there exists some index k N, such that an , n > k. bn Using (3) we get the inequalities a1 + a2 + + an a1 + + ak + (bk+1 + bk+2 + . . . bn ), n > k. (4) (3)
If we denote for simplicity the sums a1 + + an by An and b1 + + bn by Bn , the above inequality reads: An Ak + (Bn Bk ), n > k, so dividing by Bn we get An Ak Bk + . Bn Bn (5)
Since Bn , by xing k and taking lim sup in (5), we get lim supn (An /Bn ) . In other words, we obtained the inequality lim sup
n
a1 + + an , L, b1 + + bn a1 + + an L. b1 + + bn 1
which in turn forces lim sup
n
Remark. An equivalent formulation of the above Theorem is as follows: If (yn ) n=1 is a strictly increasing sequence with limn yn = , then for any sequence (xn ) , the n=1 following inequalities hold: lim sup xn xn xn1 lim sup ; n yn n yn yn1 xn xn xn1 lim inf lim inf . n yn n yn yn1 xn xn1 yn yn1 lim
(6) (7)
In particular, if the sequence
has a limit, then
n=1
xn xn xn1 = lim . n yn n yn yn1 Indeed (assuming all the yn s are positive, which happens anyway for n large enough), if we consider the sequences (an ) and (bn ) , dened by a1 = x1 , b1 = y1 , and an = xn xn1 , n=1 n=1 bn = yn yn1 , n 2, then everything is clear, since xn = a1 + + an and yn = b1 + + bn . The Stolz-Cesaro Theorem has numerous applications in Calculus. Below are three of the most signicant ones. Additive Cesaros Theorem. For any sequence (an ) R one has the inequalin=1 ties: lim sup a1 + a2 + + an lim sup an ; n n n a1 + a2 + + an lim inf lim inf an . n n n (8) (9)
In particular, if the sequence (an ) has a limit, then n=1 a1 + a2 + + an = lim an . n n n lim Proof. Particular case of Stolz-Cesaro Theorem with bn = 1. Remark. An equivalent formulation of the above Theorem (proven using the alternative version of Stolz-Cesaro Theorem) is as follows: For any sequence (xn ) , the following n=1 inequalities hold: lim sup xn lim sup(xn xn1 ); n n n xn lim inf lim inf (xn xn1 ). n n n (10) (11)
In particular, if the sequence (xn xn1 ) has a limit, then n=1 xn = lim (xn xn1 ). n n n lim 2
Multiplicative Cesaros Theorem. For any sequence of positive numbers (an ) n=1 one has the inequalities: lim sup n a1 a2 an lim sup an ; (12) n n (13) lim inf n a1 a2 an lim inf an .
n n
In particular, if the sequence (an ) has a limit, then n=1 lim n a1 a2 an = lim an .
n n
Proof. Let bn = ln an , so that
a1 a2 . . . an = exp
b1 + + bn . Everything then follows n
from the Additive Cesaro Theorem. Remark. An equivalent formulation of the above Theorem (proven using the alternative version of Additive Cesaro Theorem) is as follows: For any sequence of positive numbers (xn ) , the following inequalities hold: n=1 xn lim sup n xn lim sup ; (14) n n xn1 xn . (15) lim inf n xn lim inf n n xn1 In particular, if the sequence (xn /xn1 ) has a limit, then n=1 xn lim n xn = lim . n n xn1 LHopitals Rule. Suppose f and g are dierentiable on some interval that has A as f (x) = L. Then: an accumualtion point, limxA g(x) = , and limxA g (x) f (x) = L. xA g(x) lim (Comment: The notation limxA can include any kind of limit: honest, one-sided, or A = . It is also implicitly assumed that g (x) = 0 for x near A. Nothing is assumed about limxA f (x), not even its existence!) Proof. By a simple change-of-variable argument, it suces to consider only the case A = . Since g (x) = 0 on some interval (, ) and limx g(x) = , it follows that g is strictly increasing on (, ). By a standard argument, it suces to show that ( ) Whenever (tn ) is a strictly increasing sequence in (, ), with limn tn = , it n=1 follows that: f (tn ) lim = L. (16) n g(tn ) 3
To prove ( ) x a sequence (tn ) as above, and let us consider the sequences xn = f (tn ) n=1 and yn = g(tn ). Using the Lagranges Mean Value Theorem, we know that for every n 2, there exists some sn in (tn1 , tn ), such that f (tn ) f (tn1 ) f (sn ) = . g(tn ) g(tn1 ) g (sn ) Since (sn ) is obviously increasing, with limn sn = , it follows that limn n=1 L. Going back to (17), we now have xn xn1 = L, n yn yn1 lim with (yn ) strictly increasing (because g is strictly increasing) and limn yn = . By n=1 xn = L, which is the (alternative version of) Stolz-Cesaro Theorem, it follows that limn yn precisely the desired conclusion (16). (17) f (sn ) = g (sn )