Proving convergent sequence theorem.

Question

When $n$ approach to infinity prove that if

$$ \lim(a_{n+1}-a_n))= 0,$$ then $a_n$ is convergent.

I can prove the converse of this theorem is true but I can't prove this one. I know that since

$$ \lim_{n\to \infty}(a_{n+1}-a_n))= 0, $$ we got for all $ε>0$, there exists an $N$ such that for all $n>N$, $|a_{n+1} -a_n| <ε$.

I also know that $|a_{n+1} -a_n| ≥ |a_{n+1}|-|a_n|$ that means $|a_{n+1}|-|a_n|<ε$. Now I don't know what to do next.

If $a_n=\sum_1^n1/k$, then $\lim(a_{n+1}-a_n)=0$, but $a_n$ is not convergent. — Gerry Myerson, Sep 15 '13 at 11:28

score 2 · Accepted Answer · answered Sep 15 '13 at 12:14

2

Take $$a_n = \sum_{k=1}^n\frac 1k.$$ Clearly, $\lim_{n\to\infty}|a_{n}-a_{n-1}| = \lim_{n\to\infty}\frac 1n = 0$, yet $$\lim_{n\to\infty}a_n=+\infty.$$

answered Sep 15 '13 at 12:14

TZakrevskiy

22,980

@GerryMyerson I apologise, must have missed your comment. – TZakrevskiy Sep 16 '13 at 11:52

score 1 · Answer 2 · answered Sep 15 '13 at 11:43

1

Check this sequence

$$ a_n = \ln(n). $$

answered Sep 15 '13 at 11:43

Mhenni Benghorbal

47,431

Proving convergent sequence theorem.

2 Answers2