Show that the inequality is valid for infinite terms of a sequence 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 $n \in \mathbb{N}$.
 A: 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. 
A: 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)
