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.

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    $\begingroup$ It's hard to understand: what is the sequence, anyway? $\endgroup$ – DonAntonio Apr 28 '17 at 8:37
  • $\begingroup$ "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? $\endgroup$ – Crostul Apr 28 '17 at 8:37
  • $\begingroup$ @DonAntonio The sequence is not given. $\endgroup$ – Avatrin Apr 28 '17 at 8:38
  • $\begingroup$ @Crostul The assumption is that the sequence diverges... I'll edit the question to make that clear. $\endgroup$ – Avatrin Apr 28 '17 at 8:38
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    $\begingroup$ @Avatrin You are aware that "$a_n$ diverges" is not the same as "$a_n\to+\infty$"? $\endgroup$ – Hagen von Eitzen 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 $\lim(a_{n+1}-a_n/2)=0-0/2=0$.)

  • $\begingroup$ [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). $\endgroup$ – Especially Lime Apr 28 '17 at 9:12
  • $\begingroup$ Yeah, you are right. I actually deleted my comment before you posted since I realized the test was useless for my purpose. So, yeah, with this clarification, my question has been answered completely. $\endgroup$ – Avatrin Apr 28 '17 at 9:14

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