How to prove this statement using Cauchy's statement?

Question

If $b_n>0$ is a sequence which implies that for every $\epsilon>0$ exists a certain $N$ so that for $m>n>N$ the expression $\sum_{k=n}^m b_k$< $\epsilon$ is true.It is also known that for every natural $n$ the equation $|a_{n+1} - a_n|\le b_n$ I need to prove that $a_n$ converges.How to prove the last expression using Cauchy's statements?

Answer 1

We can prove that $\sum_{n=m}^\infty a_n$ converges if and only if, for every real num­ ber $\epsilon > 0$, there exists an integer $N\ge m$ such that $$\left|\sum_{n=p}^q a_n\right|\le\epsilon\text{ for all } p, q\ge N.$$

To prove that statement we can use the fact that a sequence is convergent if and only if it is a cauchy sequence.

Proof. The direct. Let $S_M=\sum_{n=m}^M a_n$ the partial sum of the series. By hypothesis $(S_M)_{M=n}^\infty$ is convergent, so it is a Cauchy sequence. Then there exists a $N\ge m$ such that $$\left|S_q-S_p\right|\le\epsilon$$ for all $p,q\ge N$. Suppose $p\le q$ without of generality, so we know that $$\left|S_q-S_p\right|=\left|\sum_{n=m}^q a_n-\sum_{n=m}^p a_n\right|=\left|\sum_{n=p+1}^q a_n\right|.$$ We can continue the proof from here.

Basically the sequence $(b_n)_{n=m}^\infty$ satisfies this statement. So the series is convergent and hence it is cauchy.

Answer 2

Let an $\epsilon>0$ be given. We have to produce an $N$ such that $|a_m-a_n|<\epsilon$ whenever $m>n>N$.

Now $a_m-a_n=\sum_{k=n}^{m-1}(a_{k+1}-a_k)$ and therefore $$|a_m-a_n|\leq\sum_{k=n}^{m-1} |a_{k+1}-a_k|\leq\sum_{k=n}^{m-1} b_k\ .$$ By assumption on the $b_k>0$ there is an $N$ such that the right hand side is $<\epsilon$ whenever $n>N$, so that we are done.