# How come $\lim_{n\rightarrow \infty} \frac{a_n}{a_{n-1}}$ be different than 1?

There is a theorem that "$\forall_{n}: a_n>0 ~and~ \lim_{n\rightarrow \infty} \frac{a_n}{a_{n-1}}=L \Rightarrow \lim_{n\rightarrow \infty} \sqrt[\leftroot{-2}\uproot{2}n]{a_n}=L$.

Does the left hand side of the statement also implies that $a_n$ does not converges to a finite limit? (since if $a_n$ has a limit $L$ then $a_{n-1}$ has the exact same limit $L$. Now, a series $c_n=\frac{a_n}{b_n}$ has a limit $L_c=\frac{L_a}{L_b}$. Thus, $\lim_{n\rightarrow \infty} \frac{a_n}{a_{n-1}}=\frac{L}{L}=1$). Then the remaining cases are $a_n$ converges to $\infty$ or not at all.

• if $|L| < 1$ then a_n converges to $0.$ if $L > 1, a_n$ does not converge. If $L = 1,$ then $a_n$ may converge to a finite limit, but it may not. – Doug M Aug 12 '16 at 17:50
• If $a_n \to L$, then $$\lim_{n \to \infty}\frac{a_n}{a_{n-1}} = \frac{L}{L} = 1$$ is only true if $L \neq 0$. – Bungo Aug 12 '16 at 18:01

## 3 Answers

Basically the left hand side is the exact same as the ratio test. So in other words if $L < 1$ then it will converge. If $L > 1$ it won't converge. If $L=1$ then you need another test.

What you are confusing is that $\lim_{n \rightarrow \infty} c_n = \frac{L_a}{L_b}$ if the limits of $a_n$ and $b_n$ exist and $L_b \neq 0$.

So basically any sequence that has a limit that goes to zero or infinity means that the ratio test doesn't have to go to one.

Edit: Even though the ratio test is for series it is still true that if the series converges then the sequence must converge (to zero) as well.

Hint: define $a_n=1/2^n, a_n/a_{n-1}=1/2$ but the sequences converges towards 0

• The title suggests that the question is asking why it is so. – GoodDeeds Aug 12 '16 at 17:46
• please read the entire question in particular the bold characters – Tsemo Aristide Aug 12 '16 at 17:47

The problem in your argument lies in the fact that when the sequence converges, $\lim_{n\rightarrow\infty}a_n=0$. So, for, $$L=\lim_{n\rightarrow\infty}\frac{a_n}{a_{n-1}}\ne\frac{\lim_{n\rightarrow\infty}a_n}{\lim_{n\rightarrow\infty}a_{n-1}}$$ as we have $\lim_{n\rightarrow\infty}a_{n-1}=0$.

So, you don't get it in the form $\frac{L}{L}=1$.