I am solving the following problem:
a.) Let $(a_n)$ be a sequence such that $|a_{n+1} −a_n| < C^n$ for all $n$ for some constant $0 < C < 1$ . Prove that $(a_n)$ is a Cauchy sequence, therefore is convergent.
This is what I have:
$C^n + C^n+1+…+C^m, S=C^n+C^{n+1}+…+C^m, CS=C^{n+1}+C^{n+2}+…+C^{m+1}(1-0)=C^n-C^m+1, S=C^n-C^m+1$
b.) Let $(a_n)$ be a sequence such that $|a_{n+1} −a_n| <\frac{1}{n}$ for all $n$. Is it true that such $(a_n)$ is always convergent?
Im not sure about what I have for part a, and I haven’t begun part b.)
Thanks.
Consider the series $$\sum_{i=1}^\infty |a_{i+i} - a_i|$$
Since the terms in this series satisfy $|a_{n+1} - a_n| <C^n$ for each $n$, we can say that $$\sum_{n=1}^\infty |a_{n+1} - a_n| < \sum_{n=1}^\infty C^n$$
The series on the right is a geometric series with common ratio $C \in (0,1)$. Hence, the geometric series converges. Furthermore, since the series on the left is bounded below by $0$ and consists of terms that are non-negative real numbers, the series on the left must converge. This implies that the "tail" of the series on the left gets arbitrarily small. In other words, if we fix $\epsilon > 0$, there exists $M > 0$ such that
$$\sum_{n=M}^\infty |a_{n+1} - a_n| < \epsilon$$
Then, we note that if we pick two numbers $n>m > M$, we have $$|a_n - a_m| = |(a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + \ldots + (a_{m+1} - a_m)| \leq \sum_{i=m}^n |a_{i+1}-a_i|\leq \sum_{i=M}^\infty|a_{i+1} - a_i|$$
where the first inequality that appears follows by repeated applications of the triangle inequality. The sum on the very right we know to be smaller than $\epsilon$, so it follows $|a_n - a_m| < \epsilon$. Hence, our sequence is Cauchy.
For part (b), we need to notice how crucial the fact was that we had a geometric series in the proof of part (a). If we simply let $|a_{n+1} - a_n| < \frac{1}{n}$, we may not necessarily have had a convergent series since $$\sum_{n=1}^\infty \frac{1}{n} = \infty$$
If you can construct a sequence $(a_n)$ satisfying $|a_{n+1} - a_n| < \frac{1}{n}$ but for which $$\sum_{n=1}^\infty |a_{n+1} - a_n|$$ does not converge, you may very well be in business for a counterexample to part (b). In fact, you could let $$a_n = \sum_{i=1}^n \frac{1}{i}$$
You would just have to show this satisfies the desired inequality and it does not converge.