Show that the inequality is valid for infinite terms of a sequence

Question

This question comes from a Brazilian book of real analysis, which is "Introdução a Análise" (Introduction to Analysis) of Antonio Caminha. The problem is:

Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of positive real numbers. Show that the inequality

$$ 1 + a_n > 2^{1/n}a_{n-1} $$

is true for infinity $n \in \mathbb{N}$.

Answer 1

Suppose the inequality does not hold for infinitely many $n \in N$. Then, $ \forall$ except finitely many $n \in N$,

the reverse inequality holds, i.e., $1 + a_n \leq 2^{1/n}a_{n-1} $ -------(1)

Taking limsup on both sides, and by the positivity of the $a_n$, we get,

$ 1+ \limsup a_n \leq \limsup2^{1/n}a_{n-1}$ --- (2)

Case 1) $ \limsup a_{n} < \infty $ Let $\limsup a_n = a$ Then by (2), $ 1 + a \leq a $ A contradiction. Hence the result holds.

Case 2) $ \limsup a_n = \infty$ Then by (1), for all except finitely many $n \in N$, we have $ 1+ {a_n} \leq 2^{1/n}a_{n-1}$

(say this happens $\forall n \geq N_0$ )

Multiplying by $2^{1/{n+1}} $ on both sides, we get, $2^{1/{n+1}} + 2^{1/{n+1}}a_n \leq 2^{ 1/{n+1}} •2^{1/n}a_{n-1} \forall n \geq N_0$

Then again using the inequality (1), we get,

$ 2^{1/n+1} + 1 + a_{n+1} \leq 2^{ \frac{2n+1}{n(n+1)}}a_{n-1}$

$ \Rightarrow 1+ a_{n+1} \leq 2^{ \frac{2n+1}{n(n+1)}}a_{n-1}$

Continuing in this way, we obtain , $ 1+ a_{n+1} \leq 2^ {\frac{f(n)}{g(n)}}a_{N_0}$ $ \forall n \geq N_0$ where $f(n)$ & $g(n)$ are functions such that $\lim_{n \rightarrow\infty} \frac{f(n)}{g(n)} =0 $ ------(3) Hence,

$ a_{n+1} \leq 2^ {\frac{f(n)}{g(n)}} a_{N_0} + C $

$\forall n \in N$

( where the constant C is chosen according to the finitely many terms $a_n $ where $ 1\leq n \leq N_0$ By (3), since any convergent sequence is bounded, We have,

$a_n \leq M $ $ \forall n \in N$ for some constant

$ M >0$

Contradiction to assumption that $\limsup a_n = \infty$ Hence proved.

Answer 2

My attempt (other way to solve)

Ps: this technique is like the Gronwall inequality for sequences

Suppose that the initial inequality holds for only a finite number of coefficients, then for $n\geq n_0$, $$a_{n+1} \leq 2^{\frac{1}{n}}a_n-1 $$ multiply the both sides by $\prod\limits^{n}_{k=1}2^{\frac{1}{k}}$, we have $$\frac{a_{n+1}}{\prod\limits^{n}_{k=1}2^{\frac{1}{k}}} \leq \frac{2^{\frac{1}{n}}a_n}{\prod\limits^{n}_{k=1}2^{\frac{1}{k}}}-\frac{1}{\prod\limits^{n}_{k=1}2^{\frac{1}{k}}} $$ define $h(s)=\frac{a_{s}}{\prod\limits^{s-1}_{k=1}2^{\frac{1}{k}}} $, then $$h(s+1) \leq h(s)- \frac{1}{\prod\limits^{s}_{k=1}2^{\frac{1}{k}}}=$$ so $$h(s+1)- h(s) \leq -\frac{1}{\prod\limits^{s}_{k=1}2^{\frac{1}{k}}} $$ apply the sum $\sum\limits^{n-1}_{s=n_0}$ in both sides, the first is telescopic $$h(n)\leq h(n_0) - \sum\limits^{n-1}_{s=n_0}\frac{1}{2^{H_s}} $$ where $H_s=\sum\limits^{s}_{k=1}\frac{1}{k}$ the harmonic number. Using the defition of $h(n)$ we have $$a_{n} \leq 2^{H_n}\left( h(n_0)- \sum\limits^{n-1}_{s=n_0}\frac{1}{2^{H_s}}\right), $$ but $\sum\limits^{n-1}_{s=n_0}\frac{1}{2^{H_s}}$ diverges, then $a_n$ is negative for $n$ sufficiently large. (This divergence we can see by the comparison of $H_s$ with $\ln (s)$, $H_s<1+\ln(s)$ and cauchy condensation test)