I'm trying to prove the ratio test for sequences. Here's what I got:

If $ \lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n} = L < 1 $ and $ a_n>0 \;\ \forall n $ then $ a_n $ is bounded below by $0$. Also there's $N$ so that forall $n>N$, $a_{n+1}<a_n $. Therefore, the sequence is decreasing and bounded below so it must converge.

Now, according to the test, $ \lim \limits_{n \to \infty} a_n = 0 $. Why is that?


1 Answer 1


If we note that the limit statement is formally that for any $\epsilon>0$ we may select $N$ so that for $n>N$ we have

$$\left|{a_{n+1}\over a_n}-L\right|<\epsilon$$

but then we see, by multiplying both sides by $a_n>0$, that

$$\left|a_{n+1}-La_n\right|<\epsilon a_n$$

then we have $a_{n+1}-La_n<\epsilon a_n\implies a_{n+1}<(L+\epsilon)a_n$

choose $0<\epsilon <(1-L)$--this is positive since $0<L<1$. Then $L+\epsilon <1$ and we see then that


and as $m\to\infty$ we have $a_{n+m}\to 0$ by the squeeze theorem.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.