# Prove that if $\lim_{n\to\infty}\frac{x_{n+1}}{x_n} = L < 1$ then $\lim_{n\to\infty} x_n = 0$

Let $(x_n)$ be a sequence with $x_n > 0$ for all $n \in \mathbb{N}$. I would like a hint on how to prove that if $\lim_{n\to\infty}\frac{x_{n+1}}{x_n} = L < 1$ then $\lim_{n\to\infty} x_n = 0$.

Here is what i have tried. There exist a $N$ such that $|\frac{x_{n+1}}{x_n} - L| < \varepsilon$ for all $n \geq N$. From this i must conclude that for some $M$ we have $|x_n|<\varepsilon$ for $n \geq M$. Can anyone point me in the right direction? I suspect i need some smart trick.

Thanks.

How about this reverse argument: since $\lim_{n\to \infty}x_{n+1}/x_n=L<1$, then, by the ratio test, the series $$\sum_{n=1}^\infty x_n < \infty$$ That gives $$\lim_{n\to \infty}x_n=0$$

• Reverse argument, rather than direct, since the ratio test is based on the result this question asks to prove. – Did Apr 19 '15 at 17:49
• @Did, that's what I tought, although we can prove the ratio test whitout this result. I put it this way, since the OP did not specify if he already know the ratio test. – Alonso Delfín Apr 19 '15 at 18:00
• @Did, for example by proving that the series is absolutely convergent. See en.m.wikipedia.org/wiki/Ratio_test for the outline of the proof – Alonso Delfín Apr 19 '15 at 20:02
• How exactly does the proof of the ratio test use $x_n\to 0$? – zhw. Apr 19 '15 at 20:16
• Well, in my opinion Did has a point... but I don't agree that the ratio test depends on this, as he says. It "comes off" in the proof of the ratio test, but the fact by itself is of no use in the test: knowing it makes no difference for the proof. It is the same as saying that we use the fact that $a_n \rightarrow 0$ to prove the cauchy criterion for series. Nevertheless, I agree that using a strictly more powerful result that "spits out" what you want to prove is not advisable, even more because OP is probably getting introduced to analysis. – Aloizio Macedo Apr 20 '15 at 1:07

Suppose the limit exists, and is different than $0$. Let's call it $x$. Then, $\lim \frac{x_{n+1}}{x_n}=\frac{x}{x}=1<1$. We have arrived at a contradition.

Therefore, if it exists, it is $0$.

Pick $L<\gamma<1$. Therefore, there exists $N$ such that for all $n>N$, $x_{n+1}<\gamma x_n<x_n$. We then have the sequence is decreasing after $N$. But if it is decreasing and bounded, it converges.

We can summarize the argument as follows:

Okay, let's suppose the limit exists. Then it must be $0$. But the sequence is eventually monotonic and bounded. Therefore, it exists, and is $0$.

Since $L<1$ so choose $\epsilon$ such that $L+\epsilon, L-\epsilon <1$. Now applying the definition of limit to this $\epsilon$ and then applying recursion, you can see that $(L-\epsilon)^mx_N < x_{N+m}<(L+\epsilon)^m$. Now use the Sandwich Theorem.

If $\lim_{n\to\infty}\frac{x_{n+1}}{x_n}=L$, then for some $N$, $0<\frac{x_{n+1}}{x_n}<q\forall n>N$ where $q$ is some number between $L$ and $1$. (You can set $q$ as $L+\varepsilon$ where $\epsilon$ is small enough so that $L+\varepsilon<1$).

Now, notice that for all $n>N$, $0<x_{N+p}=\frac{x_{N+p}}{x_{N+p-1}}\cdot\frac{x_{N+p-1}}{x_{N+p-2}}\cdot\ldots\cdot\frac{x_{N+1}}{x_N}\cdot c<c\cdot q^{p}$

Where $c$ is a constant equal to $\frac{x_N}{x_{N-1}}\cdot\ldots\cdot\frac{x_2}{x_1}$

Taking the limit of the right side and noting that $|q|<1$, we find that the limit of the sequence $x_n$ is equal to $0$.

QED.

Use inequality, 2nd form: $$\Biggl\lvert\frac{x_{n+1}}{x_n}-L\Biggr\rvert\le\Biggl\lvert\frac{x_{n+1}}{x_n}-L\Biggr\rvert$$ If the latter is $<\varepsilon$ for all $n\ge N$, you deduce that: $$0<\frac{x_{n+1}}{x_n}<L+\varepsilon,$$ hence if $L+\varepsilon\le k <1$, we have $\,0<x_n<\dfrac{k^N}{x_N}k^{n}$, i. e. if $n$ is large enough , $x_n$ is bounded from above by a geometric sequence that converges to $0$.

We can find an $N$ so that if $n\ge N$, we have $\frac{x_{n+1}}{x_n}\le\frac{L+1}2$. This means that for $n\ge N$, $x_n$ is a decreasing sequence bounded below by $0$. Therefore, $\lim\limits_{n\to\infty}x_n$ exists and is $\ge0$.

Suppose that $$\lim_{n\to\infty}x_n=x_\infty$$ Taking the limit of $$x_{n+1}\le\frac{L+1}2x_n$$ yields $$x_\infty\le\frac{L+1}2x_\infty$$ Subtracting the right side from both sides and multiplying by $\frac2{1-L}$, which is positive, gives $$x_\infty\le0$$ Therefore, $x_\infty=0$.