The proof I am following is based on Bernoulli inequality, which says that $(1+x)^n\ge 1+nx$, for $x>-1$ and $n \in N$. It is possible to show by Bernoulli inequality that sequence $x_n$ is increasing for $n>|x|$. Then comes the bounded part, which is not fully explained, actually only the case where $x=1$ is explicitly proved. It is based on the fact that sequences $(1+1/n)^n$ and $(1-1/n)^n$ are increasing as special cases of sequence $x_n$ and $(1+1/n)(1-1/n)\le 1$, which implies that $(1+1/n)^n\le(1-1/n)^{-n}\le(1-1/2)^{-2}=4$. Can the argument be generalized just by doing the same thing with increasing sequences $(1+x/n)^n$ and $(1-x/n)^n$? So the bound would be something like $4^{|x|}$.
Proving that sequence $x_n=(1+x/ n)^n$ is increasing and bounded for n>|x| and therefore the $\lim_{n \to \infty}x_n$ exists
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real-analysis
1 Answers
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Yes, it appears to work.
Let $n$ be the smallest natural number so that $\dfrac x n < 1$. Then using your suggested estimates find for all $m \ge n$:
$\left(1+\frac x m\right)^m \le \left(1-\frac x m\right)^{-m} \le \left(1-\frac x n\right)^{-n}$
While it is a bit hard to imagine what this bound precisely is (it depends on $x$ nontrivially), it is at least constant if $x$ is fixed, so we deduce convergence by the Monotone Convergence Theorem.