Does the Series Sum of Differences Converge Given a Contractive Condition?

[evinda]

Hello! (Wave)

Let $0< \theta<1$ and a sequence $(a_n)$ for which it holds that

$$|a_{n+2}-a_{n+1}| \leq \theta |a_{n+1}-a_{n}|, n=1,2, \dots$$

We have already shown that $(a_n)$ converges. Could you give me a hint how we could also show that $\sum_{n=1}^{\infty} (a_{n+1}-a_n)$ converges?

 

[I like Serena]

Hello! (Wave)

Let $0< \theta<1$ and a sequence $(a_n)$ for which it holds that

$$|a_{n+2}-a_{n+1}| \leq \theta |a_{n+1}-a_{n}|, n=1,2, \dots$$

We have already shown that $(a_n)$ converges. Could you give me a hint how we could also show that $\sum_{n=1}^{\infty} (a_{n+1}-a_n)$ converges?

Hey evinda!

That's a telescoping sum isn't it?
That is:
$$\sum_{n=1}^{\infty} (a_{n+1}-a_n)
= \lim_{n\to\infty} \sum_{k=1}^{n} (a_{k+1}-a_k)
= \lim_{n\to\infty} \big((a_{n+1} - a_n) + (a_{n} - a_{n-1}) + \ldots + (a_2 - a_1)\big)
= \lim_{n\to\infty} (a_{n+1} - a_1)
$$
(Thinking)

 

[math771]

Hi there! Sure, here's a hint for showing that the series $\sum_{n=1}^{\infty} (a_{n+1}-a_n)$ converges: First, use the given inequality to show that the sequence $(a_n)$ is Cauchy. Then, use the Cauchy criterion for series to show that the series $\sum_{n=1}^{\infty} (a_{n+1}-a_n)$ converges. Let me know if you need any further clarification or help with the proof. Good luck!