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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?

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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 $$ b_n > L - \frac{L - \ell}{2} = \frac{L + \ell}{2} > \frac{\ell + \ell}{2} = \ell. $$

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  • $\begingroup$ 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. $\endgroup$ – James Apr 17 at 14:02
  • $\begingroup$ @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 $\lambda_0$... $\endgroup$ – gt6989b Apr 17 at 14:53
  • $\begingroup$ 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 $\mu\in [0,1)$? $\endgroup$ – James Apr 17 at 15:50
  • $\begingroup$ @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 $\endgroup$ – gt6989b Apr 17 at 18:31
  • $\begingroup$ Thank you for your help. By the way, this is a direct consequence of the ratio test! :) $\endgroup$ – James Apr 17 at 23:15

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