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

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evinda
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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?
 
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evinda said:
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)
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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