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I'm trying to find $\lim_{n \to \infty}{\left(1+\frac{1}{\sqrt n}\right)^n}$. It looks pretty easy, so I think there's probably something simple I'm missing, because I can't seem to figure out where to even start. I've tried some basic algebraic manipulation but it doesn't seem to help much. I know the sequence diverges to infinity from typing it into WolframAlpha, but I don't know how to show that it's unbounded.

I'm trying to do it without using L'Hospital's rule or Bernoulli's inequality. Other Calc 1 methods are fair game.

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    $\begingroup$ Just use the good old $\exp/\log$-trick and taylor expansion... $\endgroup$ – TheGeekGreek Mar 30 '18 at 16:35
  • $\begingroup$ Do you know what $\displaystyle\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n$ is? $\endgroup$ – Dustan Levenstein Mar 30 '18 at 16:35
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    $\begingroup$ Or simply $$\left(1+\frac1{\sqrt n}\right)^{\sqrt n}\geqslant2$$ $\endgroup$ – Did Mar 30 '18 at 16:36
  • $\begingroup$ Write as $((1 + \frac{1}{\sqrt{n}})^{\sqrt{n}})^{\sqrt{n}}$ $\endgroup$ – jim Mar 30 '18 at 16:36
  • $\begingroup$ possible duplicate :math.stackexchange.com/q/1681772/156150 $\endgroup$ – zeraoulia rafik Mar 30 '18 at 16:42
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Here a way without using Bernoulli directly but using the binomial formula only: $\left(1+\frac{1}{\sqrt n}\right)^n= 1 + \binom{n}{1}\frac{1}{\sqrt n} + \binom{n}{2}\left(\frac{1}{\sqrt n}\right)^2 + \cdots + \binom{n}{n}\left(\frac{1}{\sqrt n}\right)^n \stackrel{n>1}{>} 1 + \binom{n}{1}\frac{1}{\sqrt n} = 1+\sqrt{n}$

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    $\begingroup$ If use of Bernoulli's Inequality is prohibited, then use of the binomial theorem is also likely to be prohibited. $\endgroup$ – Mark Viola Mar 30 '18 at 21:10
  • $\begingroup$ I'm allowed to use the binomial theorem, actually. Thank you all for your answers :) I'm going to accept this one as it's the simplest to understand given the tools I have to work with but having different perspectives is always a great thing. $\endgroup$ – Cassiterite Mar 31 '18 at 16:10
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    $\begingroup$ @Cassiterite Bernoulli's Inequality is actually quite a bit more elementary than the binomial theorem. In fact, the first two terms of the binomial theorem for $(1+x)^n$ are $1$ and $nx$. $\endgroup$ – Mark Viola Mar 31 '18 at 17:27
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Bernoulli's Inequality gives

$$\left(1+\frac1{\sqrt n}\right)^n\ge1+\sqrt n\to \infty$$

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  • $\begingroup$ That's nice and simple, unfortunately I can't use Bernoulli's inequality either. Should have mentioned it in the question, sorry, I've updated it $\endgroup$ – Cassiterite Mar 30 '18 at 17:00
  • $\begingroup$ Very nice and direct way! $\endgroup$ – user Mar 30 '18 at 19:13
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    $\begingroup$ @gimusi Thank you! Much appreciated. $\endgroup$ – Mark Viola Mar 30 '18 at 21:11
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    $\begingroup$ @Cassiterite Are you permitted to use the inequalities $\log(x)\ge \frac{x-1}{x}$ and $e^x\ge 1+x$? If so, then we have $$\left(1+\frac1{\sqrt n}\right)^n=e^{n\log\left(1+\frac1{\sqrt n}\right)}\ge e^{\frac{n}{1+\sqrt n}}\ge 1+\frac{n}{1+\sqrt n}$$ $\endgroup$ – Mark Viola Mar 30 '18 at 21:16
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As $n\to\infty$, $$\left(1+\frac1{\sqrt n}\right)^{\sqrt n}\to e$$ and $e^{\sqrt n}$ is divergent since $e>1$.

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Note that

$$\left(1+\frac{1}{\sqrt n}\right)^n=\left[\left(1+\frac{1}{\sqrt n}\right)^{\sqrt n}\right]^{\frac n {\sqrt n}}\ge2^{\frac n {\sqrt n}}\to \infty$$

or as an alternative

$$\left(1+\frac{1}{\sqrt n}\right)^n=e^{n\log\left(1+\frac{1}{\sqrt n}\right)}\sim e^{\frac n {\sqrt n}-\frac12}\to \infty$$

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