For a sequence, how should we check that the distance between two consecutive terms $a_{n+2}$ and $a_{n+1}$ is less than $a_{n+1}$ and $a_n$?

Question

Background

It can be proved that for $ 0 < \alpha <1$, if $\{a_n\}$ is a sequence which satisfies

$$ |a_{n+2} - a_{n+1}| \leq \alpha |a_{n+1} - a_n|$$

Then $a_n$ is a Cauchy sequence.

Question

I have already proved that the statement in background is indeed true. But when I try to use this fact in practice, I find it very difficult to demonstrate this property for a specific sequence; in fact, this is particularly difficult for sequence that appears in the denominator in a recurrence relation. For example, for this sequence,

$$ b_{n+1} = \frac{1}{1+b_n}$$, where $b_1 =1$

and for this sequence,

$$c_{n+1} = \frac{3+2c_n}{3+c_n}$$, where $c_1 =1$

, I have tried convoluted methods to demonstrate that it fulfils the condition described above. My attempt is one of induction. For example:

My attempt

By induction, for $n=1$ $$|b_2 - b_1| = |\frac{1}{1+1} - 1| = \frac{1}{2}$$ $$|b_3 - b_2| = |\frac{1}{\frac{3}{2}}|- \frac{1}{2} = \frac{1}{6} < \frac{1}{2}$$ Assume that for $n=k$, $$ |b_{k+2} - b_{k+1}| \leq \alpha |b_{k+1} - b_k|$$ ...

From then on I get into complicated algebra since I tried to express $b_{k+3}$ in terms of $b_k$, the large expression eventually leads to some inequality involving higher powers of $b_k$. How should I approach such recurrent sequence with itself appearing in the denominator?

Answer 1

In my opinion, that 'background' fact is not a very useful tool; it works in only some cases and even then it can be much harder to use it than to not use it. It is often far easier to systematically prove the convergence to the limit directly than to prove that something is merely a Cauchy sequence. $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

---

Example 1

If the sequence converges to some real $r$ then it is easy to check that $r > 0$ and $r = \lfrac1{1+r}$ (and I shall leave you to find $r$). Let $p_n = b_n-r$ for every $n \in \mathbb{N}_{>0}$. Then $p_{n+1} = \lfrac1{1+r+p_n}-r$ $= \lfrac{r}{1/r+p_n} - r$ $= -\lfrac{r}{1/r+p_n}·c_n$ for every $n \in \mathbb{N}_{>0}$. Now all you have to do is to show that $0 \le \lfrac{r}{1/r+p_n} < r$ for every $n \in \mathbb{N}_{>0}$, which can be done by induction since $1/r > 1.5$ and $|p_1| < 0.5$. Using that, you can obviously prove directly that $p_n \to 0$ exponentially as $n \to \infty$. Note that the initial assumption of convergence is not used in the proof, but merely used to tell us how to define $r$.

Example 2

Exactly the same trick works here. Let real $s > 0$ such that $s = \lfrac{3+2s}{3+s}$, and $q_n = c_n-s$ for every $n \in \mathbb{N}_{>0}$. Then $q_{n+1} = \lfrac{3+2s+2q_n}{3+s+q_n} - s$ $= \lfrac{2-s}{3+s+q_n}·q_n$. We can prove that $0 \le \lfrac{s-2}{3+s+q_n} < \lfrac{s-2}3$, because $s+q_n = c_n > 0$, and hence easily prove that $q_n \to 0$ exponentially as $n \to \infty$.

---

This technique is powerful enough that we can easily prove the quadratic rate of convergence of the Newton-Raphson approximation method with explicit conditions (not just a weak claim that it has such convergence rate in some open interval around the root).

---

Anyway, here is a nice problem which can be solved using either Cauchy convergence or the monotone convergence theorem.

Does the infinite sum $\lfrac{1}{4} + \lfrac{1·3}{4·6} + \lfrac{1·3·5}{4·6·8} + \lfrac{1·3·5·7}{4·6·8·10} + \cdots$ converge?

Hint:

Prove that $(1·3·5·7)^2 \le 1·(2·4·6)^2·7$, giving $\lfrac{1·3·5·7}{4·6·8·10} \le \lfrac{\sqrt{7}}{2·8·10}$, and likewise for the other terms.

Answer 2

Let $f(x)=1/(1+x)$. Then, by the Mean Value Theorem, $$ b_{n+1}-b_n=f(b_n)-f(b_{n-1})=f'(c)(b_n-b_{n-1}) $$ for some $c$ between $b_n$ and $b_{n+1}$. It will be enough to prove that $|f'(c)|\le\alpha<1$. First of all, it is easy to see that $1/2\le b_b\le1$. Then also $1/2\le c\le1$, and $$ |f'(c)|=\frac{1}{(1+c)^2}\le\frac{1}{(1+1/2)^2}=\frac49<1. $$ For the other sequence, use $f(x)=(3+2\,x)/(3+x)$.