# Stolz–Cesàro theorem

Stolz–Cesàro theorem In mathematics, the Stolz–Cesàro theorem is a criterion for proving the convergence of a sequence. The theorem is named after mathematicians Otto Stolz and Ernesto Cesàro, who stated and proved it for the first time.

The Stolz–Cesàro theorem can be viewed as a generalization of the Cesàro mean, but also as a l'Hôpital's rule for sequences.

Contents 1 Statement of the theorem for the ∙/∞ case 2 Statement of the theorem for the 0/0 case 3 Proofs 3.1 Proof of the theorem for the ∙/∞ case 3.2 Proof of the theorem for the 0/0 case 4 Applications and examples 4.1 Arithmetic mean 4.2 Geometric mean 4.3 Examples 4.3.1 Example 1 4.3.2 Example 2 5 History 6 The general form 6.1 Statement 6.2 Proof 6.2.1 Proof of the equivalent statement 6.2.2 Proof of the original statement 7 References 8 External links 9 Notes Statement of the theorem for the ∙/∞ case Let {displaystyle (a_{n})_{ngeq 1}} and {displaystyle (b_{n})_{ngeq 1}} be two sequences of real numbers. Assume that {displaystyle (b_{n})_{ngeq 1}} is a strictly monotone and divergent sequence (i.e. strictly increasing and approaching {displaystyle +infty } , or strictly decreasing and approaching {displaystyle -infty } ) and the following limit exists: {displaystyle lim _{nto infty }{frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}=l. } Then, the limit {displaystyle lim _{nto infty }{frac {a_{n}}{b_{n}}}=l. } Statement of the theorem for the 0/0 case Let {displaystyle (a_{n})_{ngeq 1}} and {displaystyle (b_{n})_{ngeq 1}} be two sequences of real numbers. Assume now that {displaystyle (a_{n})to 0} and {displaystyle (b_{n})to 0} while {displaystyle (b_{n})_{ngeq 1}} is strictly decreasing. If {displaystyle lim _{nto infty }{frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}=l, } then {displaystyle lim _{nto infty }{frac {a_{n}}{b_{n}}}=l. } [1] Proofs Proof of the theorem for the ∙/∞ case Case 1: suppose {displaystyle (b_{n})} strictly increasing and divergent to {displaystyle +infty } , and {displaystyle -infty 0} there exists {displaystyle nu >0} such that {displaystyle forall n>nu } {displaystyle left|,{frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}-l,right|<{frac {epsilon }{2}},} which is to say {displaystyle l-epsilon /2<{frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}nu .} Since {displaystyle (b_{n})} is strictly increasing, {displaystyle b_{n+1}-b_{n}>0} , and the following holds {displaystyle (l-epsilon /2)(b_{n+1}-b_{n})nu } .

Next we notice that {displaystyle a_{n}=[(a_{n}-a_{n-1})+dots +(a_{nu +2}-a_{nu +1})]+a_{nu +1}} thus, by applying the above inequality to each of the terms in the square brackets, we obtain {displaystyle {begin{aligned}&(l-epsilon /2)(b_{n}-b_{nu +1})+a_{nu +1}=(l-epsilon /2)[(b_{n}-b_{n-1})+dots +(b_{nu +2}-b_{nu +1})]+a_{nu +1}0} such that {displaystyle b_{n}>0} for all {displaystyle n>n_{0}} , and we can divide the two inequalities by {displaystyle b_{n}} for all {displaystyle n>max{nu ,n_{0}}} {displaystyle (l-epsilon /2)+{frac {a_{nu +1}-b_{nu +1}(l-epsilon /2)}{b_{n}}}<{frac {a_{n}}{b_{n}}}<(l+epsilon /2)+{frac {a_{nu +1}-b_{nu +1}(l+epsilon /2)}{b_{n}}}.} The two sequences (which are only defined for {displaystyle n>n_{0}} as there could be an {displaystyle Nleq n_{0}} such that {displaystyle b_{N}=0} ) {displaystyle c_{n}^{pm }:={frac {a_{nu +1}-b_{nu +1}(lpm epsilon /2)}{b_{n}}}} are infinitesimal since {displaystyle b_{n}to +infty } and the numerator is a constant number, hence for all {displaystyle epsilon /2>0} there exists {displaystyle n_{pm }>n_{0}>0} , such that {displaystyle {begin{aligned}&|c_{n}^{+}|n_{+},\&|c_{n}^{-}|n_{-},end{aligned}}} therefore {displaystyle l-epsilon max lbrace nu ,n_{pm }rbrace =:N>0,} which concludes the proof. The case with {displaystyle (b_{n})} strictly decreasing and divergent to {displaystyle -infty } , and {displaystyle l0} there exists {displaystyle nu >0} such that for all {displaystyle n>nu } {displaystyle {frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}>2M.} Again, by applying the above inequality to each of the terms inside the square brackets we obtain {displaystyle a_{n}>2M(b_{n}-b_{nu +1})+a_{nu +1},quad forall n>nu ,} and {displaystyle {frac {a_{n}}{b_{n}}}>2M+{frac {a_{nu +1}-2Mb_{nu +1}}{b_{n}}},quad forall n>max{nu ,n_{0}}.} The sequence {displaystyle (c_{n})_{n>n_{0}}} defined by {displaystyle c_{n}:={frac {a_{nu +1}-2Mb_{nu +1}}{b_{n}}}} is infinitesimal, thus {displaystyle forall M>0,exists {bar {n}}>n_{0}>0{text{ such that }}-M{bar {n}},} combining this inequality with the previous one we conclude {displaystyle {frac {a_{n}}{b_{n}}}>2M+c_{n}>M,quad forall n>max{nu ,{bar {n}}}=:N.} The proofs of the other cases with {displaystyle (b_{n})} strictly increasing or decreasing and approaching {displaystyle +infty } or {displaystyle -infty } respectively and {displaystyle l=pm infty } all proceed in this same way.

Proof of the theorem for the 0/0 case Case 1: we first consider the case with {displaystyle l0} , we can write {displaystyle a_{n}=(a_{n}-a_{n+1})+dots +(a_{n+nu -1}-a_{n+nu })+a_{n+nu },} and for any {displaystyle epsilon /2>0,} {displaystyle exists n_{0}} such that for all {displaystyle n>n_{0}} we have {displaystyle {begin{aligned}&(l-epsilon /2)(b_{n}-b_{n+nu })+a_{n+nu }=(l-epsilon /2)[(b_{n}-b_{n+1})+dots +(b_{n+nu -1}-b_{n+nu })]+a_{n+nu }0} there are {displaystyle nu _{pm }>0} such that {displaystyle {begin{aligned}&|c_{nu }^{+}|nu _{+},\&|c_{nu }^{-}|nu _{-},end{aligned}}} thus, choosing {displaystyle nu } appropriately (which is to say, taking the limit with respect to {displaystyle nu } ) we obtain {displaystyle l-epsilon n_{0}} which concludes the proof.

Case 2: we assume {displaystyle l=+infty } and {displaystyle (b_{n})} strictly decreasing. For all {displaystyle 2M>0} there exists {displaystyle n_{0}>0} such that for all {displaystyle n>n_{0},} {displaystyle {frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}>2Mimplies a_{n}-a_{n+1}>2M(b_{n}-b_{n+1}).} Therefore, for each {displaystyle nu >0,} {displaystyle {frac {a_{n}}{b_{n}}}>2M+{frac {a_{n+nu }-2Mb_{n+nu }}{b_{n}}},quad forall n>n_{0}.} The sequence {displaystyle c_{nu }:={frac {a_{n+nu }-2Mb_{n+nu }}{b_{n}}}} converges to {displaystyle 0} (keeping {displaystyle n} fixed). Hence {displaystyle forall M>0,~exists {bar {nu }}>0} such that {displaystyle -M{bar {nu }},} and, choosing {displaystyle nu } conveniently, we conclude the proof {displaystyle {frac {a_{n}}{b_{n}}}>2M+c_{nu }>M,quad forall n>n_{0}.} Applications and examples The theorem concerning the {displaystyle cdot /infty } case has a few notable consequences which are useful in the computation of limits.

Arithmetic mean Let {displaystyle (x_{n})} be a sequence of real numbers which converges to {displaystyle l} , define {displaystyle a_{n}:=sum _{m=1}^{n}x_{m}=x_{1}+dots +x_{n},quad b_{n}:=n} then {displaystyle (b_{n})} is strictly increasing and diverges to {displaystyle +infty } . We compute {displaystyle lim _{nto infty }{frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}=lim _{nto infty }x_{n+1}=lim _{nto infty }x_{n}=l} therefore {displaystyle lim _{nto infty }{frac {x_{1}+dots +x_{n}}{n}}=lim _{nto infty }x_{n}.} Given any sequence {displaystyle (x_{n})_{ngeq 1}} of real numbers, suppose that {displaystyle lim _{nto infty }x_{n}} exists (finite or infinite), then {displaystyle lim _{nto infty }{frac {x_{1}+dots +x_{n}}{n}}=lim _{nto infty }x_{n}.} Geometric mean Let {displaystyle (x_{n})} be a sequence of positive real numbers converging to {displaystyle l} and define {displaystyle a_{n}:=log(x_{1}cdots x_{n}),quad b_{n}:=n,} again we compute {displaystyle lim _{nto infty }{frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}=lim _{nto infty }log {Big (}{frac {x_{1}cdots x_{n+1}}{x_{1}cdots x_{n}}}{Big )}=lim _{nto infty }log(x_{n+1})=lim _{nto infty }log(x_{n})=log(l),} where we used the fact that the logarithm is continuous. Thus {displaystyle lim _{nto infty }{frac {log(x_{1}cdots x_{n})}{n}}=lim _{nto infty }log {Big (}(x_{1}cdots x_{n})^{frac {1}{n}}{Big )}=log(l),} since the logarithm is both continuous and injective we can conclude that {displaystyle lim _{nto infty }{sqrt[{n}]{x_{1}cdots x_{n}}}=lim _{nto infty }x_{n}} .

Given any sequence {displaystyle (x_{n})_{ngeq 1}} of (strictly) positive real numbers, suppose that {displaystyle lim _{nto infty }x_{n}} exists (finite or infinite), then {displaystyle lim _{nto infty }{sqrt[{n}]{x_{1}cdots x_{n}}}=lim _{nto infty }x_{n}.} Suppose we are given a sequence {displaystyle (y_{n})_{ngeq 1}} and we are asked to compute {displaystyle lim _{nto infty }{sqrt[{n}]{y_{n}}},} defining {displaystyle y_{0}=1} and {displaystyle x_{n}=y_{n}/y_{n-1}} we obtain {displaystyle lim _{nto infty }{sqrt[{n}]{x_{1}dots x_{n}}}=lim _{nto infty }{sqrt[{n}]{frac {y_{1}dots y_{n}}{y_{0}cdot y_{1}dots y_{n-1}}}}=lim _{nto infty }{sqrt[{n}]{y_{n}}},} if we apply the property above {displaystyle lim _{nto infty }{sqrt[{n}]{y_{n}}}=lim _{nto infty }x_{n}=lim _{nto infty }{frac {y_{n}}{y_{n-1}}}.} This last form is usually the most useful to compute limits Given any sequence {displaystyle (y_{n})_{ngeq 1}} of (strictly) positive real numbers, suppose that {displaystyle lim _{nto infty }{frac {y_{n+1}}{y_{n}}}} exists (finite or infinite), then {displaystyle lim _{nto infty }{sqrt[{n}]{y_{n}}}=lim _{nto infty }{frac {y_{n+1}}{y_{n}}}.} Examples Example 1 {displaystyle lim _{nto infty }{sqrt[{n}]{n}}=lim _{nto infty }{frac {n+1}{n}}=1.} Example 2 {displaystyle {begin{aligned}lim _{nto infty }{frac {sqrt[{n}]{n!}}{n}}&=lim _{nto infty }{frac {(n+1)!(n^{n})}{n!(n+1)^{n+1}}}\&=lim _{nto infty }{frac {n^{n}}{(n+1)^{n}}}=lim _{nto infty }{frac {1}{(1+{frac {1}{n}})^{n}}}={frac {1}{e}}end{aligned}}} where we used the representation of {displaystyle e} as the limit of a sequence.

History The ∞/∞ case is stated and proved on pages 173—175 of Stolz's 1885 book and also on page 54 of Cesàro's 1888 article.

It appears as Problem 70 in Pólya and Szegő (1925).

The general form Statement The general form of the Stolz–Cesàro theorem is the following:[2] If {displaystyle (a_{n})_{ngeq 1}} and {displaystyle (b_{n})_{ngeq 1}} are two sequences such that {displaystyle (b_{n})_{ngeq 1}} is monotone and unbounded, then: {displaystyle liminf _{nto infty }{frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}leq liminf _{nto infty }{frac {a_{n}}{b_{n}}}leq limsup _{nto infty }{frac {a_{n}}{b_{n}}}leq limsup _{nto infty }{frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}.} Proof Instead of proving the previous statement, we shall prove a slightly different one; first we introduce a notation: let {displaystyle (a_{n})_{ngeq 1}} be any sequence, its partial sum will be denoted by {displaystyle A_{n}:=sum _{mgeq 1}^{n}a_{m}} . The equivalent statement we shall prove is: Let {displaystyle (a_{n})_{ngeq 1},(b_{n})_{geq 1}} be any two sequences of real numbers such that {displaystyle b_{n}>0,quad forall nin {mathbb {Z} }_{>0}} , {displaystyle lim _{nto infty }B_{n}=+infty } , then {displaystyle liminf _{nto infty }{frac {a_{n}}{b_{n}}}leq liminf _{nto infty }{frac {A_{n}}{B_{n}}}leq limsup _{nto infty }{frac {A_{n}}{B_{n}}}leq limsup _{nto infty }{frac {a_{n}}{b_{n}}}.} Proof of the equivalent statement First we notice that: {displaystyle liminf _{nto infty }{frac {A_{n}}{B_{n}}}leq limsup _{nto infty }{frac {A_{n}}{B_{n}}}} holds by definition of limit superior and limit inferior; {displaystyle liminf _{nto infty }{frac {a_{n}}{b_{n}}}leq liminf _{nto infty }{frac {A_{n}}{B_{n}}}} holds if and only if {displaystyle limsup _{nto infty }{frac {A_{n}}{B_{n}}}leq limsup _{nto infty }{frac {a_{n}}{b_{n}}}} because {displaystyle liminf _{nto infty }x_{n}=-limsup _{nto infty }(-x_{n})} for any sequence {displaystyle (x_{n})_{ngeq 1}} .

Therefore we need only to show that {displaystyle limsup _{nto infty }{frac {A_{n}}{B_{n}}}leq limsup _{nto infty }{frac {a_{n}}{b_{n}}}} . If {displaystyle L:=limsup _{nto infty }{frac {a_{n}}{b_{n}}}=+infty } there is nothing to prove, hence we can assume {displaystyle L<+infty } (it can be either finite or {displaystyle -infty } ). By definition of {displaystyle limsup } , for all {displaystyle l>L} there is a natural number {displaystyle nu >0} such that {displaystyle {frac {a_{n}}{b_{n}}}nu .} We can use this inequality so as to write {displaystyle A_{n}=A_{nu }+a_{nu +1}+dots +a_{n}nu ,} Because {displaystyle b_{n}>0} , we also have {displaystyle B_{n}>0} and we can divide by {displaystyle B_{n}} to get {displaystyle {frac {A_{n}}{B_{n}}}<{frac {A_{nu }-lB_{nu }}{B_{n}}}+l,quad forall n>nu .} Since {displaystyle B_{n}to +infty } as {displaystyle nto +infty } , the sequence {displaystyle {frac {A_{nu }-lB_{nu }}{B_{n}}}to 0{text{ as }}nto +infty {text{ (keeping }}nu {text{ fixed)}},} and we obtain {displaystyle limsup _{nto infty }{frac {A_{n}}{B_{n}}}leq l,quad forall l>L,} By definition of least upper bound, this precisely means that {displaystyle limsup _{nto infty }{frac {A_{n}}{B_{n}}}leq L=limsup _{nto infty }{frac {a_{n}}{b_{n}}},} and we are done.

Proof of the original statement Now, take {displaystyle (a_{n}),(b_{n})} as in the statement of the general form of the Stolz-Cesàro theorem and define {displaystyle alpha _{1}=a_{1},alpha _{k}=a_{k}-a_{k-1},,forall k>1quad beta _{1}=b_{1},beta _{k}=b_{k}-b_{k-1},forall k>1} since {displaystyle (b_{n})} is strictly monotone (we can assume strictly increasing for example), {displaystyle beta _{n}>0} for all {displaystyle n} and since {displaystyle b_{n}to +infty } also {displaystyle mathrm {B} _{n}=b_{1}+(b_{2}-b_{1})+dots +(b_{n}-b_{n-1})=b_{n}to +infty } , thus we can apply the theorem we have just proved to {displaystyle (alpha _{n}),(beta _{n})} (and their partial sums {displaystyle (mathrm {A} _{n}),(mathrm {B} _{n})} ) {displaystyle limsup _{nto infty }{frac {a_{n}}{b_{n}}}=limsup _{nto infty }{frac {mathrm {A} _{n}}{mathrm {B} _{n}}}leq limsup _{nto infty }{frac {alpha _{n}}{beta _{n}}}=limsup _{nto infty }{frac {a_{n}-a_{n-1}}{b_{n}-b_{n-1}}},} which is exactly what we wanted to prove.

References Mureşan, Marian (2008), A Concrete Approach to Classical Analysis, Berlin: Springer, pp. 85–88, ISBN 978-0-387-78932-3. Stolz, Otto (1885), Vorlesungen über allgemeine Arithmetik: nach den Neueren Ansichten, Leipzig: Teubners, pp. 173–175. Cesàro, Ernesto (1888), "Sur la convergence des séries", Nouvelles annales de mathématiques, Series 3, 7: 49–59. Pólya, George; Szegő, Gábor (1925), Aufgaben und Lehrsätze aus der Analysis, vol. I, Berlin: Springer. A. D. R. Choudary, Constantin Niculescu: Real Analysis on Intervals. Springer, 2014, ISBN 9788132221487, pp. 59-62 J. Marshall Ash, Allan Berele, Stefan Catoiu: Plausible and Genuine Extensions of L’Hospital's Rule. Mathematics Magazine, Vol. 85, No. 1 (February 2012), pp. 52–60 (JSTOR) External links l'Hôpital's rule and Stolz-Cesàro theorem at imomath.com Proof of Stolz–Cesàro theorem at PlanetMath. Notes ^ Choudary, A. D. R.; Niculescu, Constantin (2014). Real Analysis on Intervals. Springer India. pp. 59–60. ISBN 978-81-322-2147-0. ^ l'Hôpital's rule and Stolz-Cesàro theorem at imomath.com This article incorporates material from Stolz-Cesaro theorem on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.

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