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I am working on a proof as follows:

Let $k > 1$ and $(a_n)_{n \in \Bbb{N}}$ be a sequence where $a_n > 0$ for all $n \in \Bbb{N}$.

$$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = k$$

Show that $\lim_{n \to \infty} a_n = +\infty$

I have ascertained that when n is large enough, $\frac{a_{n+1}}{a_n} > 1$ and that $a_{n+1} > a_n$ so $(a_n)$ will be strictly monotone increasing.

However, this is not strong enough to show the sequence tends to infinity.

I feel like I understand the notion of why it tends to infinity, the strict inequality meaning it won't converge however I'm not sure how to prove it.

Any help will be greatly appreciated.

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  • $\begingroup$ Show that $\dfrac{a_{n+1}}{a_n} \geqslant \sqrt{k}$ for all large enough $n$. $\endgroup$ Nov 19, 2016 at 16:20

4 Answers 4

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Choose an $\epsilon > 0$ such that $1+\epsilon < k$. We know that there exists an $N$ such that $$\frac{a_{n+1}}{a_n}>1+\epsilon$$ for all $n\ge N$. Thus we have that $a_{n+1}>(1+\epsilon)a_n$. Then applying this recursively we have that $$a_{N+k}>(1+\epsilon)^ka_N$$ And we see that $a_N>0$ and $(1+\epsilon)^k\to\infty$ so $a_{N+k}\to\infty$.

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  • $\begingroup$ Will, I understand. I would encourage you to be more precise on the choice of $\epsilon$ here. See my post for details. ;-)) -Mark $\endgroup$
    – Mark Viola
    Nov 19, 2016 at 17:07
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Herein, we present a standard approach that relies on the definition of the limit. To that end we proceed.


If $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=k>1$, then for all $\epsilon>0$, there exists a number $N(\epsilon)$ such that whenever $n>N(\epsilon)$,

$$k-\epsilon<\frac{a_{n+1}}{a_n}<k+\epsilon$$

Take $\epsilon=\frac{k-1}{2}$. Then, for $n>N\left(\frac{k-1}{2}\right)$, we see that

$$a_{n+1}>\left(\frac12+\frac k2\right)a_n$$

where $\left(\frac12+\frac k2\right)>1$.

Proceeding recursively, we find that

$$a_{n+m}>\left(\frac12+\frac k2\right)^ma_n$$

Letting $m\to \infty$, we obtain the coveted limit

$$\lim_{n\to \infty}a_n=\infty$$

And we are done!

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Hint

there exists $a>1$ and $N\geq 0$ s.t

for $n\geq N$,

$$\frac{u_n}{u_{n-1}}\geq a$$ $$\frac{u_{n-1}}{u_{n-2}}\geq a$$

.

.

$$\frac{u_{N+1}}{u_N}\geq a$$

use telescoping and geometric sequence.

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lim{a_(n+1)}=lim {a_(n+1)/a_n}*{a_n}=lim{a_(n+1)/a_n}*lim{a_n}. This only works if lim a_n is infinity.

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