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How does one go about calculating the limit of the sequence $$a_n = \left(1 + \dfrac{1}{n^2} \right)^n.$$ I understand the cases of $ b_n = \left(\alpha + \dfrac{\beta}{n}\right)^n, $ in terms of $e$, but I am missing the "trick" with regards to $a_n$.

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Since $\left(1+\dfrac{1}{n^{2}}\right)^{n^{2}}\rightarrow e$, then $n^{2}\log\left(1+\dfrac{1}{n^{2}}\right)\rightarrow\log e=1$. Now \begin{align*} n\log\left(1+\dfrac{1}{n^{2}}\right)=\dfrac{1}{n}\cdot n^{2}\log\left(1+\dfrac{1}{n^{2}}\right)\rightarrow 0\cdot 1=0, \end{align*} so \begin{align*} \left(1+\dfrac{1}{n^{2}}\right)^{n}\rightarrow e^{0}=1. \end{align*}

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Note that

$$1\le\left(1 + \dfrac{1}{n^2} \right)^n=\left[\left(1 + \dfrac{1}{n^2} \right)^{n^2}\right]^\frac1n\le e^\frac1n$$

thus for squeeze theorem

$$\left(1 + \dfrac{1}{n^2} \right)^n\to 1$$

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    $\begingroup$ Gimusi. Very nice, you do not forget to bound below:))) $\endgroup$ – Peter Szilas Jan 13 '18 at 19:04
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Hint:

$$\lim_{n\to\infty}\left(1+\frac1{n^2}\right)^{n^2}=e$$

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As long as you know the binomial theorem, you don't need to know anything about $e$:

$$\begin{align} 1\le\left(1+{1\over n^2}\right)^n &=1+{n\over n^2}+{{n\choose2}\over n^4}+\cdots+{{n\choose n}\over n^{2n}}\\ &\le1+{1\over n}+{n^2\over n^4}+\cdots+{n^n\over n^{2n}}\\ &\le1+{1\over n}+{1\over n^2}+\cdots+{1\over n^n}+\cdots\\ &={1\over1-{1\over n}}\to1 \end{align}$$

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  • $\begingroup$ In the last equality wouldn't it be as follow? $$={{1-\left(\frac1n\right)^{n+1}}\over1-{1\over n}}\to1$$ $\endgroup$ – user Jan 13 '18 at 19:18
  • $\begingroup$ @gimusi, the line above the last line extended the finite sum to the infinite geometric series. $\endgroup$ – Barry Cipra Jan 14 '18 at 5:10
  • $\begingroup$ Oh yes of course! Thanks $\endgroup$ – user Jan 14 '18 at 8:49
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I thought it might be of interest to present a way forward that relies only on Bernoulli's Theorem and the squeeze theorem. To that end we proceed.


This is as simple as $1$, $2$, and $3$.

First, we note that

$$\left(1+\frac{1}{n^2}\right)\le \frac1{\left(1-\frac1{n^2}\right)}\tag 1$$


Second, applying Bernoulli's inequality ($(1+x)^n\ge 1+x$ for $x>-1$) to both sides of $(1)$ reveals

$$1+\frac1n\le \left(1+\frac{1}{n^2}\right)^n\le \frac{1}{1-\frac1n}\tag2$$


Third, and finally, applying the squeeze theorem to $(1)$ yields the coveted limit

$$\lim_{n\to \infty}\left(1+\frac{1}{n^2}\right)^n=1$$


And we are done!

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You have obtains $$ \lim_{m\rightarrow \infty}\left(1+\frac{1}{m}\right)^m =e.$$

In your case, you have that $m=n^2$, then $$ \lim_{n\rightarrow \infty}\left(1+\frac{1}{n^2}\right)^n=\lim_{n\rightarrow \infty}\left(1+\frac{1}{n^2} \right)^{n^2\cdot\frac{1}{n} }=\left[ \lim_{n\rightarrow \infty}\left(1+\frac{1}{n^2} \right)^{n^2 } \right]^{\frac{1}{n}}=e^{\lim_{n\rightarrow \infty}\frac{1}{n}}=e^0=1.$$

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