# Proof about the convergence of a sequence

In another question here, we were given an identity based off a sequence:

$\lim_{n \rightarrow \infty}a_{n+1}-\frac{a_n}{2} = 0$

I tried to show that the sequence converges. However, I am very uncertain whether my proof is correct or not.

If we assume $a_n$ diverges:

$\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} -\frac{1}{2}= \lim_{n \rightarrow \infty}\frac{a_{n+1}-\frac{a_n}{2}}{a_n} \rightarrow \frac{0}{\infty} = 0$

This contradicts the assumption that the sequence diverges due to the ratio test, and it must thus be false.

• It's hard to understand: what is the sequence, anyway? Apr 28 '17 at 8:37
• "This contradicts our assumption due to the ratio test": what assumption does it contradict? This is exactly the assumption we have. Anyway, what sequence are you talking about, and what do you apply the ratio test to? Apr 28 '17 at 8:37
• @DonAntonio The sequence is not given. Apr 28 '17 at 8:38
• @Crostul The assumption is that the sequence diverges... I'll edit the question to make that clear. Apr 28 '17 at 8:38
• @Avatrin You are aware that "$a_n$ diverges" is not the same as "$a_n\to+\infty$"? Apr 28 '17 at 8:48

I'm not quite clear what you are trying to prove. Is it that $a_n$ is a convergent series, i.e. $\sum_n a_n$ is finite, or that it is a convergent sequence, i.e. $\lim_{n\to\infty}a_n$ exists and is finite?
For the first, this is not true and the fallacy is that you've assumed that if it isn't a convergent series then $a_n\to\infty$. For example, setting $a_n=\frac1n$ satisfies the original condition, but doesn't give a convergent series.
For the second, this is true, but the ratio test is not relevant (it is a test for convergence of the series, not the sequence). Your proof also doesn't work because it only shows that $a_n\not\to\infty$, but the sequence could conceivably fail to converge because there is no limit, rather than because the limit is infinite.
In fact we must have $\lim_{n\to\infty}a_n=0$. To see this, note that for any $\delta>0$ we have some $N$ such that $|a_{n+1}|\leq|a_n|/2+\delta$ for all $n\geq N$. Thus $|a_{N+2}|\leq |a_{N+1}|/2+\delta\leq|a_N|/4+\delta+\delta/2$. Continuing in this manner, you can show that $|a_{N+k}|< 2^{-k}|a_N|+2\delta$ for every $k$, and so for all $k$ sufficiently large you get $|a_{N+k}|<3\delta$.
(In fact a sequence satisfies your original condition if and only if it tends to $0$; if $\lim a_n=0$ then $\lim(a_{n+1}-a_n/2)=0-0/2=0$.)
• [reply to a deleted comment] Yes, it's true that passing the ratio test implies that the sequence tends to $0$ (because that must be true whenever the series converges). But it's no help for testing if a sequence is convergent to something other than $0$, which is why I thought you might be trying to show the series was convergent (if I'd known the question asked you to show specifically that it converged to $0$, I wouldn't have thought that). Apr 28 '17 at 9:12