# The limit of the difference of two consecutive sequence members is equal to $0$. Can we conclude that the sequence itself has a limit?

Let $a_n$ be an infinite sequence. The limit of the difference of two consecutive members is equal to $0$. Can we conclude that the sequence itself has a limit?

My attempt: We have $$\lim_{n\rightarrow\infty}{a_n} - \lim_{n\rightarrow\infty}{a_{n-1}} = 0$$ since as $n$ approaches infinity $a_{n-1}$ gets arbitrarily close to $a_n$ the sequence cannot diverge or be bounded but have no limit.

Is my proof correct and how would I be able to formalize the last sentence? Thanks

• When you write $\lim_{n\to\infty} a_n - \lim_{n\to\infty} a_{n-1} = 0$, you seem to be implicitly assuming that both limits exist, and are equal. But you are trying to show that $\lim_{n\to \infty} a_n$ exists. Your argument is circular. – Xander Henderson Sep 12 '17 at 16:03
• Sorry. you're right. – mtheorylord Sep 12 '17 at 16:04

While I can't entirely follow your proof, it seems to be assuming the existence of $\lim_{n\to\infty}a_n$, which is what you're trying to prove or disprove.

Here's an example which shows that you cannot conclude that $\lim_{n\to\infty}a_n$ exists: Let $a_n=1+\frac{1}{2}+\dots+\frac{1}{n}$. Then $a_{n+1}-a_n=\frac{1}{n+1}\to0$, but the sequence $\{a_n\}$ diverges.

Your proof is not correct because you assume what you want to prove.

Now take the sequence $a_n=\sqrt{n}$

Then $a_{n+1}-a_n=\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}} \to 0$

But $a_n \to +\infty$

Also this sequences gives you a reason to see why your proof does not work.

And another counter-example $a_n=\ln{n}$, we have $\lim\limits_{n \rightarrow \infty} a_n=+\infty$ (still an acceptable notation), but $$a_{n+1}-a_n=\ln{(n+1)}-\ln{n}=\ln{\left(\frac{n+1}{n}\right)}=\ln{\left(1+\frac{1}{n}\right)}\rightarrow\ln{1}=0, n\rightarrow\infty$$

NO. $\lim_{n\to \infty} (a_{n+1}-a_n)=0$ is NOT the same sentence as $\lim_{n\to \infty}a_{n+1}-\lim_{n\to \infty}a_n=0.$

Example. Let $a_n=\sum_{j=1}^n(1/j).$ Then $a_{n+1}-a_n=1/(n+1)\to 0$ as $n\to \infty$ but the sequence $(a_n)_{n\in \mathbb N}$ has no upper bound. $$a_4>a_2+2(1/4)=2.$$ $$a_8>a_4+4(1/8)>2+4(1/8)=5/2.$$ $$a_{16}>a_8+8(1/16)>5/2+8(1/16)=3.$$ Et cetera.