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For $x\in\mathbb{R}$ and $n\in\mathbb{N}$, define the real sequence: $$a_n:=(1+x/n)^n$$ Now prove that there exists an $\alpha\in\mathbb{R}$ (just prove that $\alpha$ exists, no need to find the value) such that: $$\forall\epsilon>0,\exists N\in\mathbb{N}:\forall n\ge N,|a_n-\alpha|<\epsilon$$

This should be done strictly with basic algebra; no logarithms, no Taylor series, no derivatives, and $e$ shouldn't appear anywhere in the proof. I suspect that the Bernoulli inequality will probably be useful. Considering the limitations, is there a concise way to show this?

EllipticalInitial
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  • Hint: First show that it holds for any $x$ from the existence of $\lim_{n\to \infty}(1+1/n)^n.$ – DanielWainfleet Mar 13 '18 at 02:58
  • Bernoulli is the key here. You should first prove that the limit exists for $x=1$ (this is famous and should be known to you). Use algebraic manipulation to show that if the limit exists for $x=1$ it exists for all rational $x$. – Paramanand Singh Mar 13 '18 at 03:48
  • And by the way you do some some analysis and not just algebraic manipulation. In particular you need the theorem that any bounded monotone sequence is convergent. – Paramanand Singh Mar 13 '18 at 03:50
  • See related https://math.stackexchange.com/a/2404327/72031 extension to irrational $x$ is based on the fact that the sequence is eventually monotonically increasing and if $a, b$ are rationals with $a<x<b$ then the sequence is bounded above by $e^b$ or $e^a$. – Paramanand Singh Mar 13 '18 at 03:55

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MONOTONICITY

In THIS ANSWER, I showed using Bernoulli's Inequality that $\left(1+\frac xn\right)^n$ is monotonically increasing for $x>-n$.

BOUNDED ABOVE

Using the binomial theorem, we have for $x>-n$

$$\begin{align} 0&\le\left(1+\frac xn\right)^n\\\\ &=\left|\sum_{k=0}^n\binom{n}{k}\left(\frac xn\right)^k\right|\\\\ &=\left|\sum_{k=0}^n \prod_{\ell=1}^{k-1} \left(1-\frac{\ell}{n}\right)\frac{x^k}{k!}\right|\\\\ &\le \sum_{k=0}^n \frac{|x|^k}{k!}\\\\ &\le \sum_{k=0}^\infty \frac{|x|^k}{k!} \end{align}$$

where the ratio test guarantees that the series $\sum_{k=0}^\infty \frac{|x|^k}{k!}$ converges for every $x$.

CONCLUSION:

Inasmuch as $\left(1+\frac xn\right)^n$ is bounded above and monotonically increasing for each fixed $x$, then it converges for each fixed $x$.

And we are done!

Mark Viola
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