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.
 A: 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 $\lim(a_{n+1}-a_n/2)=0-0/2=0$.)
