# Eventually increasing or decreasing sequence of ratios

Let $$(a_n)_{n\geq 0}$$ be a positive sequence of real numbers. I have two questions about sequences that are related to log-convex and log-concave. I'll just copy the necessary part in an article (Lemma 4.1) I'm reading

Question 1:

Suppose that $$(a_{n+1}/a_n)_{n\geq 0}$$ is eventually increasing to $$\lambda\in (1,\infty]$$. Then the proof says: for any $$\lambda_0\in (1,\lambda)$$, we have $$a_{n+1}\geq \lambda_0 a_n$$ for $$n$$ sufficiently large. Why?

Question 2:

Suppose that $$(a_{n+1}/a_n)_{n\geq 0}$$ is eventually decreasing to $$\lambda\in [0,\infty)$$. Then, the proof says: if $$\lambda\in[0,1)$$, then $$\sum a_n<\infty$$. Why?

The first statement says that $$a_{n+1}/a_n \le \lambda$$ but $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lambda \in (1,\infty).$$ So if $$1 < \lambda_0 < \lambda$$, eventually (for $$n$$ large enough), we will have $$\frac{a_{n+1}}{a_n} > \lambda_0 \iff a_{n+1} > \lambda_0 a_n.$$

Here, $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lambda \in [0,1)$$ so analogously letting $$0 \le \lambda \le \mu < 1$$, for $$n$$ large enough you will get $$a_{n+1} < \mu a_n$$, and from that $$n$$ onwards, we can bound $$\sum a_n$$ by a geometric series.

UPDATE

Let $$(b_n)_{n = 0}^\infty$$ be a sequence with $$b_n \le L \in (1,\infty)$$ and assume $$\lim_n b_n \to L$$. Let $$\ell \in (1,L)$$. Prove that for $$n$$ large enough we have $$b_n > \ell$$.

Proof

Since $$\lim_n b_n \to L$$, there must exist $$N \in \mathbb{N}$$ such that $$\forall \epsilon >0$$ we have $$|b_n - L| < \epsilon$$ whenever $$n > N$$. In particular, pick $$\epsilon = (L- \ell)/2$$. Hence, there must exists $$N \in \mathbb{N}$$ such that for all $$n > N$$ we have $$|b_n - L| < \epsilon$$. Since $$b_n < L$$, we have $$|b_n - L| = L-b_n$$ and our inequality becomes $$L -b_n < \frac{L - \ell}{2}.$$ Eliminating $$L/2$$ and solving for $$b_n$$ on the LHS we get $$b_n > L - \frac{L - \ell}{2} = \frac{L + \ell}{2} > \frac{\ell + \ell}{2} = \ell.$$

• Thanks for your answer! You wrote on your first part that $\frac{a_{n+1}}{a_n}>\lambda_0$ eventually. Why is this the case? That was my question. – James Apr 17 at 14:02
• @James see the update for a formal argument. The idea is, since you are increasing to a limit, you can get as close to $\lambda$ as you like for $n$ large enough. So you can get within half the distance between $\lambda$ and $\lambda_0$... – gt6989b Apr 17 at 14:53
• Thanks for the update. It's been very helpful. I've one last question on your last part. You let $\mu\in [\lambda,1)$. Then in a similar proof (as you have written), one gets $a_{n+1}<\mu a_n$ for all large $n$. Then you wrote about a geometric series, I do not understand it. Did you mean that $\sum_{n}\mu^n a_n<\infty$ since $\mu\in [0,1)$? – James Apr 17 at 15:50
• @James if $a_{n+1} < \mu a_n$ for all $n \ge N$, then $a_{n+2} < \mu a_{n+1} < \mu^2 a_n$ and analogously $a_{n+k} < \mu^k a_n$ and you get $$\sum_{k=n+1}^\infty a_{n+k} < \sum_{k=n+1}^\infty \mu^k a_n = a_n \sum_{k=n+1}^\infty \mu^k$$ where the right hand side is now a geometric series – gt6989b Apr 17 at 18:31
• Thank you for your help. By the way, this is a direct consequence of the ratio test! :) – James Apr 17 at 23:15