Why viewing sequences as functions allows to find the equivalent?

Question

I want to know why this method works so well to find an equivalent of a sequence. The idea of this method is to consider the sequence like a function that can be derived.

Here are a few examples :

• Let $u_0>0$ and $u_{n+1}:=\ln(1+u_n)$ : we have $(u_n)_n$ decreasing and positive $(0\le\ln(1+u_n)\le u_n)$, so it converges to $0$.
  Now, let's consider this way of thinking : $u'(n) \approx u(n+1) - u(n) = \ln(1+u(n))-u(n) \approx -2^{-1}u^2(n)$ when $n$ approaches $+\infty$. We get : $(u(n)^{-1})' \approx 2^{-1}$. Using this information, let us look :
  $ (u(n)^{-1})' \approx u_{n+1}^{-1}-u_n^{-1}=\ln(1+u_n)^{-1}-u_n^{-1} = (u_n -2^{-1}u_n^2+o(u_n^2))^{-1}-u_n^{-1}$
  So $u_{n+1}^{-1}-u_n^{-1} = u_n^{-1}(1+2^{-1}u_n+o(u_n))-u_n$.
  We get by a telescoping sum: $u_n^{-1}\sim n/2$ and finally we have : $u_n\sim 2/n$.
• Let $u_0>0$ and $u_{n+1}:=u_n+u_n^{-1}$ : We can show that it goes to $+\infty$.
  We have : $u'(n)\approx u(n+1)-u(n) = u(n)^{-1}$. We get : $(u(n)^2)'=2^{-1}$.
  Now, let us look :
  $(u(n)^2)'\approx u_{n+1}^2-u_n^2=u_n^2(1+u_n^{-2})^2-u_n^2= u_n^2(1+2u_n^{-2}+o(u_n^{-2}))-u_n^2 \sim 2$.
  We get : $u_n^{2}\sim 2n$ and finally we have : $u_n\sim \sqrt{2n}$.
• Let $u_n >0$ and $\forall n\ge1, u_{n+1}:=u_n + (nu_n)^{-1}$ : We can show that it goes to $l$ where $l\in\mathbb R_*^+\cup\{+\infty\}$ .
  We have : $u'(n)\approx u(n+1)-u(n) = (nu(n))^{-1}$. We get : $(u(n)^2)'=2n^{-1}$.
  Now, let us look : $(u(n)^2)' \approx u_{n+1}^2-u_n^2=u_n^2(1+(\sqrt{n}u_n)^{-2})^2-u_n^2$
  So $u_{n+1}^2-u_n^2 = u_n^2(1+2(\sqrt{n}u_n)^{-2}+o(\sqrt{n}u_n)^{-2})-u_n^2 \sim 2n^{-1}$.
  We get : $u_n^{2}\sim 2\ln(n)$ and finally we have : $u_n\sim \sqrt{2\ln(n)}$.
• Let $(u_n)_n\in \mathbb R^\mathbb {N_*}$ and $S_n:=\sum_{i=1}^n u_i^2$ such that $u_nS_n\to 1$ when $n$ goes to $\infty$. We can show that $u_n\to0$ and $S_n\to+\infty$.
  We have : $S'(n)\approx S(n+1)-S(n) = u(n+1)^2\approx (S(n+1))^{-2} \approx (S(n))^{-2}$.
  We get : $(S(n)^3)'=3$.
  Then : $(S(n)^3)'\approx S_{n+1}^3-S_n^3 = (S_{n+1}-S_n)(S_{n+1}^2+S_nS_{n+1}+S_n^2)$.
  However : $S_{n+1}-S_n = u_{n+1}^2 \sim (S_n)^{-2}$.
  So : $S_{n+1}^3-S_n^3 \sim (S_n)^{-2}(S_{n+1}^2+S_nS_{n+1}+S_n^2)\sim 1+(S_{n+1}/S_n)+(S_{n+1}/S_n)^2 \sim 3.$
  Finally, we get : $S_n\sim (3n)^{1/3}$. So : $a_n \sim S_n^{-1} \sim (3n)^{-1/3}$

In these examples, considering the sequence as a function derived allows us to determine a equivalent. Why does it work so well ?

Answer 1

We have that $\mathbb N \subseteq \mathbb R$ therefore for any sequence

$$a_n: I\subseteq\mathbb N \to \mathbb R$$

we can consider, under some conditions, the corresponding differentiable function

$$a(x): A\subseteq\mathbb R \to \mathbb R$$

such that $\forall n \in I\subseteq A$ we have $a_n=a(n)$ and therefore, in particular, the asymptotic behaviour of the sequence can be studied using the corresponding function.

The key fact is that we are dealing with $n\to \infty$ and therefore $a(n+1)-a(n)=\frac{a(n+1)-a(n)}{(n+1)-n}$, under some circumstances, is "well" approximated by the derivative $a'(n)$. This is indeed the idea which connects Cesaro-Stolz theorem for sequences to L'Hopital's rule.

Let consider for example

• $a_n=n^m \implies (n+1)^m-n^m \approx mn^{m-1} =a'(n)$
• $a_n=\frac1n \implies \frac1{n+1}-\frac1n \approx -\frac1{n^2}=a'(n)$
• $a_n=\log n \implies \log(n+1)-\log n\approx \frac1{n}=a'(n)$

but

• $a_n=\sin n \implies \sin(n+1)-\sin n \not \approx a'(n)=\cos n$

Refer also to

• How to understand intuitively the Stolz-Cesaro Theorem for sequences?
• Derivative of a sequence