# If $\lim_{n\to\infty} (a_{n+1}-a_n)=0$ and $|a_{n+2}-a_n|<\frac{1}{2^n}$ then $(a_n)$ converges

Let $(a_n)$ be a sequence such that $\lim_{n\to\infty} (a_{n+1}-a_n)=0$ and $|a_{n+2}-a_n|<\frac{1}{2^n}$ for all $n$. I have to decide whether or not $(a_n)$ converges.

My attempt: I think it converges. Let $b_n=a_{2n}, c_n=a_{2n-1}$. Then:

$$|b_{n+1}-b_n|=|a_{2n+2}-a_{2n}|<\frac{1}{2^{2n}}$$

$$|c_{n+1}-c_n|=|a_{2n+1}-a_{2n-1}|<\frac{1}{2^{2n-1}}$$

Thus $(b_n)$ and $(c_n)$ are Cauchy (proven in another question) and converge. Because $(a_{2n}-a_{2n-1})$ is a subsequence of $(a_{n+1}-a_n)$ it also converges to $0$. Thus $$\lim_{n\to\infty} (b_n-c_n)=\lim_{n\to\infty} (a_{2n}-a_{2n-1})=0$$

or

$$\lim_{n\to\infty} b_n=\lim_{n\to\infty}c_n$$

Because the subsequences $(b_n)$ and $(c_n)$ cover the sequence $(a_n)$ and because they converge to the same point, $(a_n)$ converges.

Is it correct? What do you think?

• Absolutely correct. – Alex M. Apr 21 '15 at 13:31
• Isn't it enough to ask for $\lim_{n\to \infty} a_{n+1}-a_n=0$? – Alberto Debernardi Apr 21 '15 at 13:34
• @albertodebernardi Is it really enough? Doesn't the sequence ln n have the property that the difference between terms tends to 0, but isn't convergent? – Ilham Apr 21 '15 at 13:45
• Actually I answered a question some days ago which was very similar to what you've posted: math.stackexchange.com/questions/1239056/… . In view of the answer, I would conjecture that if $|a_n-a_{n+1}|\leq c_n$ and $c_n$ is a summable sequence, then $a_n$ is Cauchy and therefore convergent. – Alberto Debernardi Apr 21 '15 at 14:04
• @AlbertoDebernardi - if I understand correctly, the condition you presented is needed to prove that $(b_n)$ and $(c_n)$ are Cauchy and therefore convergent (which leads to the convergence of $(a_n)$), correct? And the given inequality is just a special case of your condition, right? – user233191 Apr 21 '15 at 14:43

Here you have a sufficient condition (from the discussion in the comments) for your sequence to be convergent: Suppose that $|a_n-a_{n+1}|\leq c_n$, where $\{c_n\}$ is such that $$\sum_{n=1}^{\infty}c_n<\infty.$$ We define the sequence of positive numbers $C_n=\sum_{k=1}^nc_n$, which is increasing. Since it converges, it is also Cauchy. Now we prove that $a_n$ is Cauchy: let $m> n$ $$|a_n-a_m|=|a_n-a_{n+1}+a_{n+1}-\cdots -a_m|\leq \sum_{k=n}^{m-1} |a_k-a_{k+1}|\leq \sum_{k=n}^{m-1} c_k = C_m-C_{n-1} \to0$$ as $m,n\to \infty$.
EDIT: Actually we only need the summability of the sequence $\{|a_n-a_{n+1}|\}$. Such sequences $\{a_n\}$ are called of bounded variation. So, supposing $$\sum_{n=1}^{\infty}|a_n-a_{n+1}|<\infty,$$ then it suffices to take $c_n=|a_n-a_{n+1}|$ in the previous argument.