# Convergence of a sequence

There exists a constant $c$ in $(0,1)$ such that:

$$|a_{n+2} – a_{n+1}|≤c|a_{n+1} – a_n|$$ for all $n$ in the natural numbers.

Then prove that an converges.

• Hint: $\sum_i c^i$ converges. – ForgotALot Feb 5 '17 at 17:40

Hint: Show that $(a_n)_n$ is a Cauchy sequence. Therefore:

1. show $|a_{n+2}-a_{n+1}|\leq c^{n+1}|a_1-a_0|$
2. estimate $|a_n-a_m|$ by using step 1 and dont forget about the geometric sum and series, where you have to use $0<c<1$.

You have: $$a_n=\sum_{k=1}^n(a_k-a_{k-1}) +a_0\qquad |a_k-a_{k-1}|≤c|a_{k-1}-a_{k-2}|≤...≤c^{k-1}|a_1-a_0|$$

And so $$\sum_{k=1}^n|a_k-a_{k-1}|≤\sum_{k=1}^n c^{k-1} |a_1-a_0|$$ Since $0<c<1$ the sum $\sum_k a_k-a_{k-1}$ converges absolutely, being majorised by the absolutely convergent sum $\sum_{k=1}^n c^{k-1}|a_1-a_0|$.

It follows $a_n$ converges.

• ok I kind of understand this, the only part I am confused about is how you got to c^k-1. Also why are you using series to prove a sequence converges. – Sasha Feb 5 '17 at 17:44
• $|a_k -a_{k-1}|≤c|a_{k-1}-a_{k-2}|≤c^2|a_{k-2}-a_{k-3}|≤c^3|a_{k-3}-a_{k-4}|$ and so on. If you keep applying the inequality you will eventually land at $|a_k-a_{k-1}|≤c^{k-1}|a_1-a_0|$. – s.harp Feb 5 '17 at 17:46
• The estimates show that the sequence is a cauchy sequence. Cauchy and convergent in this special case equivalent since we have an complete space. – Niklas Feb 5 '17 at 17:47
• @Niklas thats right, this cannot show convergence in $\Bbb Q$ for example. – s.harp Feb 5 '17 at 17:48
• ok I think I got it, however I am still confused why you are using series to prove something is Cauchy. – Sasha Feb 5 '17 at 17:49