MHB Proving Convergence of a Sequence with a Geometric Condition

[evinda]

Hello! (Wave)

Let $0< \theta<1$ and a sequence $(a_n)$ for which it holds that $|a_{n+2}-a_{n+1}| \leq \theta |a_{n+1}-a_n|, n=1,2, \dots$.

Could you give me a hint how we could show that $(a_n)$ converges?

 

[Euge]

Hi mathmari,

Note that by hypothesis $\lvert a_{n+2} - a_{n+1}\rvert \le \theta^n\lvert a_2 - a_1\rvert$ for all $n \ge 1$. By the triangle inequality, for all $n > m$,

$$\lvert a_n - a_m\rvert \le \lvert a_{m+1} - a_m\rvert + \lvert a_{m+2} - a_{m+1}\rvert + \cdots + \lvert a_n - a_{n-1}\rvert$$ and the latter expression is no greater than
$$(\theta^{m-1} + \theta^{m} + \cdots + \theta^{n-2})\lvert a_2 - a_1\rvert$$ Note $$\theta^{m-1} + \theta^{m} + \cdots + \theta^{n-2} \le \theta^{m-1} + \theta^{m} + \cdots = \frac{\theta^{m-1}}{1 - \theta}$$as $0 < \theta < 1$. Hence $$\lvert a_n - a_m\rvert \le \frac{\theta^{m-1}}{1-\theta}\lvert a_2 - a_1\rvert$$ for all $n > m$. Take it from here.

 

[Opalg]

Let $0< \theta<1$ and a sequence $(a_n)$ for which it holds that $|a_{n+2}-a_{n+1}| \leq \theta |a_{n+1}-a_n|, n=1,2, \dots$.

Could you give me a hint how we could show that $(a_n)$ converges?

Let $c_n = a_{n+1} - a_n$. Use the ratio test to show that the series $\sum c_n$ converges. Then notice that $\sum c_n$ is a telescoping series with partial sums of the form $a_n - a_1$.

Edit. Sorry, I didn't see that Euge had got there first.

 

[evinda]

Hi mathmari,

Note that by hypothesis $\lvert a_{n+2} - a_{n+1}\rvert \le \theta^n\lvert a_2 - a_1\rvert$ for all $n \ge 1$. By the triangle inequality, for all $n > m$,

$$\lvert a_n - a_m\rvert \le \lvert a_{m+1} - a_m\rvert + \lvert a_{m+2} - a_{m+1}\rvert + \cdots + \lvert a_n - a_{n-1}\rvert$$ and the latter expression is no greater than
$$(\theta^{m-1} + \theta^{m} + \cdots + \theta^{n-2})\lvert a_2 - a_1\rvert$$ Note $$\theta^{m-1} + \theta^{m} + \cdots + \theta^{n-2} \le \theta^{m-1} + \theta^{m} + \cdots = \frac{\theta^{m-1}}{1 - \theta}$$as $0 < \theta < 1$. Hence $$\lvert a_n - a_m\rvert \le \frac{\theta^{m-1}}{1-\theta}\lvert a_2 - a_1\rvert$$ for all $n > m$. Take it from here.

I see... thanks a lot! (Smile)