[Advanced Mathematics Notes] Stolz Theorem

Stolz's theorem can be understood as "L'Hopital's law of sequence", which reveals the relationship between the limit of the ratio of two sequences and the limit of the ratio of the difference between adjacent two terms.

Stolz's theorem

$\frac{\ast }{\infty }$type

Theorem 1 Let { an } \{a_n\}{ an​} and{ bn } \{b_n\}{ bn​} are two real number columns, where$\left\{b_n\right\}$ is strictly monotonic and tends to infinity ($+\infty$ or$-\infty$）。若极限$\underset{n\to \infty }{\lim }\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=l$ exists, then$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=l$。

Proof : Let $\underset{n\to \infty }{\lim }{b}_n=+\infty$ , and{ bn } \{b_n\}{ bn​} is strictly monotonically increasing. From$\underset{n\to \infty }{\lim }\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=l$知，$\forall \frac{}{2}>0$，$\exists N$ such that$\forall n>N$，都有$$\left|\frac{a_{n+1}-a_n}{b_{n+1}-b_n}-l\right|<\frac{}{2}$$即$$l-\frac{}{2}<\frac{a_{n+1}-a_n}{b_{n+1}-b_n}<l+\frac{}{2}$$Because { bn } \{b_n\}{ bn​} Strictly monotonically increasing, so$b_{n+1}-b_n>0$。因此$$\left(l-\frac{}{2}\right)(b_{n+1}-b_n)<a_{n+1}-a_n<\left(l+\frac{}{2}\right)(b_{n+1}-b_n)$$注意到$a_n=(a_n-a_{n-1})+(a_{n-1}+a_{n-2})+\cdots +(a_{N+2}-a_{N+1})+a_{N+1}$, then we can give $a_n$的下界：$$\begin{array}{rl}{a}_N& >\left(l-\frac{}{2}\right)(b_n-b_{n-1})+\left(l-\frac{}{2}\right)(b_{n-1}-b_{n-2})+\cdots +\left(l-\frac{}{2}\right)(b_{N+2}-b_{N+1})+a_{N+1}\\ & =\left(l-\frac{}{2}\right)(b_n-b_{N+1})+a_{N+1}\end{array}$$At the same time, an upper bound can also be given: $$a_N<\left(l+\frac{}{2}\right)(b_n-b_{N+1})+a_{N+1}$$Because $\underset{n\to \infty }{\lim }{b}_n=+\infty$，故$\exists N^{\prime }$ , such that$\forall n>N^{\prime }$ , with$b_n>0$ . (i.e.$b_n$Always positive after a term. ) Now, for n > max ⁡ { N , N ′ } n>\max\{N,N'\}n>max{ N,N′ },$a_n$Divide both sides of the upper and lower bounds by $b_n$：$$\left(l-\frac{}{2}\right)-\left(l-\frac{}{2}\right)\frac{b_{N+1}}{b_n}+\frac{a_{N+1}}{b_n}<\frac{a_n}{b_n}<\left(l+\frac{}{2}\right)-\left(l+\frac{}{2}\right)\frac{b_{N+1}}{b_n}+\frac{a_{N+1}}{b_n}$$注意$-\left(l-\frac{}{2}\right)\frac{b_{N+1}}{b_n}+\frac{a_{N+1}}{b_n}$和$-\left(l+\frac{}{2}\right)\frac{b_{N+1}}{b_n}+\frac{a_{N+1}}{b_n}$actually tends to $0$的，that is,$\forall \frac{}{2}>0$ , there exists$N_{+}>N^{\prime }$ and$N_{-}>N^{\prime }$ , so that whenn > N ∗ = max ⁡ { N + , N − } n>N^*=\max\{N_+,N_-\}n>N∗=max{ N+​,N−​}时，有$$\begin{gathered} \left|-\left(l-\frac{}{2}\right)\frac{b_{N+1}}{b_n}+\frac{a_{N+1}}{b_n}\right|<\frac{}{2} \\ \left|-\left(l+\frac{}{2}\right)\frac{b_{N+1}}{b_n}+\frac{a_{N+1}}{b_n}\right|<\frac{}{2} \end{gathered}$$故$$l-e=l-\frac{}{2}-\frac{}{2}<\frac{a_n}{b_n}<l+\frac{}{2}+\frac{}{2}=l+\epsilon$$ Therefore, we prove that:$\forall \epsilon >0$， ∃ N ∗ = max ⁡ { N + , N − } \exists N^*=\max\{N_+,N_-\} ∃N∗=max{ N+​,N−​} , such that$\forall n>N^{\ast }$，有$\left|\frac{a_n}{b_n}-l\right|<\epsilon$ ,$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=l$。

$\frac{0}{0}$type

Theorem 2 Let { an } \{a_n\}{ an​} and{ bn } \{b_n\}{ bn​} are two real numbers,$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{b}_n=0$ and$b_n$Strictly monotonically decreasing. If the limit $\underset{n\to \infty }{\lim }\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=l$ exists, then$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=l$。

Proof summary : Will $a_n$写成$a_n=(a_n-a_{n+1})+(a_{n+1}-a_{n+2})+\cdots +(a_{n-n+}-a_{n+})+a_{n+}$, then giving $a_n$The upper and lower bounds of , both sides are divided by $b_n$, notice that $a_{n+}$和$b_{n+}$Tend to $0$ can be proved.

example

1. The limit of the arithmetic mean

We want to prove that $\underset{n\to \infty }{\lim }\frac{x_1+x_2+\cdots +x_n}{n}=\underset{n\to \infty }{\lim }{x}_n$。

定义$a_n=x_1+x_2+\cdots +x_n$，$b_n=n$。则要证$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=\underset{n\to \infty }{\lim }{x}_n$. attention $a_{n+1}-a_n=x_n$，$b_{n+1}-b_n=1$ , it should now be obvious what to do.

2.

求$\underset{n\to \infty }{\lim }\left[\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right]$。

Orderan $a_n=\sqrt[n]{n!}$，$b_n=n$，则$$\underset{n\to \infty }{\lim }\left[\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right]=\underset{n\to \infty }{\lim }\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=\underset{n\to \infty }{\lim }\frac{\sqrt[n]{n!}}{n}$$New Stirling City, $n!\sim \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n$ ,if$$\underset{n\to \infty }{\lim }\frac{\sqrt[n]{n!}}{n}=\underset{n\to \infty }{\lim }\frac{{(2\pi n)}^{\frac{1}{2n}}\cdot \frac{n}{e}}{n}=\frac{1}{e}$$