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Why is $\lim\limits_{n\to\infty}(1+\frac{1}{n^2})^n = 1$?

Could someone elaborate on this? I know that $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n = e$.

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Write out the binomial expansion. You'll see why. – Frank Apr 2 '14 at 12:54
Try the substitution $n = \frac{1}{h}$ with $h \to 0$. And then exponentiate the expression i.e. $f(x) = e^{\ln f(x)}$. – Mussé Redi Apr 2 '14 at 13:00
Estimate the terms by $e^{\frac 1n}$ which is greater than each term, and has limit $1$. It is clear that the individual terms are greater than $1$. Squeeze. – Mark Bennet Apr 2 '14 at 13:01
up vote 13 down vote accepted

I know that $\lim_{n\to\infty}(1+\frac{1}{n})^n = e$

Indeed. A consequence of this statement that you know, is that there exists some finite $K\gt1$ such that, for every positive $n$, $$ 1\lt\left(1+\frac1n\right)^n\lt K, $$ (It happens that the optimal upper bound is $K=\mathrm e$, but, to solve your problem, one can forget such a refinement.) Hence, $$ 1\lt\left(1+\frac1{n^2}\right)^n=\left(\left(1+\frac1{n^2}\right)^{n^2}\right)^{1/n}\lt K^{1/n}. $$ Since $K^{1/n}\to1$ irrespectively of the value of $K$, this proves that $$ \lim_{n\to\infty}\left(1+\frac1{n^2}\right)^n=1. $$

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+1 Well explained! – Sawarnik Apr 2 '14 at 16:06

Hint: prove that $$ \log (1+u) \le u $$ enter image description here

then use the continuity of $\exp$.


$$\left(1+\frac 1{n^2} \right)^n =\exp \left(n \log\left(1+\frac 1{n^2} \right)\right) \\ 0\le n \log\left(1+\frac 1{n^2}\right) \le \frac 1n \\ 1\le \left(1+\frac 1{n^2} \right)^n \le\exp\frac 1n\to 1 $$

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How do justify $0\leq n\log\left(1 + 1/n^2\right)\leq 1/n$? – rookie Apr 2 '14 at 16:58
because $\log(1+u)\le u$, see the hint. – mookid Apr 2 '14 at 16:59

As you know that $$\lim_{n\to \infty} (1+\frac{1}{n})^n=e,$$ you also have $$\lim_{n\to \infty} (1+\frac{1}{n^2})^{n^2}=e,$$ which yields $$\lim_{n\to \infty} (1+\frac{1}{n^2})^{n}=\lim_{n\to \infty} \left((1+\frac{1}{n^2})^{n^2}\right)^{1/n}=\lim_{n\to \infty} e^{1/n}=1.$$ EDIT: The last two equalities should be seen together. It works because the interior of the parenthesis converges towards $e$ and you take the $n$-th root of some number closer and closer to $e$. When $n$ goes to infinity, this is the same as the value of the limit of $e^{1/n}$.

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I think this not rigorous. What is the justification for $$\lim_{n\to \infty} \left((1+\frac{1}{n^2})^{n^2}\right)^{1/n}=\lim_{n\to \infty} e^{1/n}$$ – Amr Apr 2 '14 at 13:10
I agree that it is a bit "unclassical". I added an edit. – Jérémy Blanc Apr 2 '14 at 13:25
A bit shaky. Look at Did's answer for a more precise proof... – mookid Apr 2 '14 at 14:37

There are some good answers here, but let me share a method that is more computational (in that it doesn't require noticing certain inequalities and using the squeeze theorem), and may help you with other similar limits.

First, one convenient way of taking exponents out of a limit expression is to take a logarithm:

$$\lim_{n \rightarrow \infty} f(n)^{g(n)} = \exp\left[\lim_{n \rightarrow \infty} g(n)\ln(f(n))\right]$$

Provided that $f(n) > 0$ (and either limit exists). In your case, this becomes:

$$\lim_{n \rightarrow \infty} (1 + \frac{1}{n^2})^{n} = \exp\left[\lim_{n \rightarrow \infty} n\ln(1 + \frac{1}{n^2})\right]$$

To evaluate the limit inside the brackets on the right, you can use l'Hospital's rule. You need to change the indeterminate form to $0/0$ and replace $n$ with a 'continuous variable' $x$:

$$\lim_{n \rightarrow \infty} n \ln(1 + \frac{1}{n^2}) = \lim_{x \rightarrow \infty} \frac{\ln(1 + \frac{1}{x^2})}{\frac{1}{x}} = \lim_{x \rightarrow \infty} \frac{(-\frac{2}{x^3})/(1 + \frac{1}{x^2})}{-\frac{1}{x^2}} = \lim_{x \rightarrow \infty} \frac{-\frac{2}{x^3}}{-\frac{1}{x^2}(1 + \frac{1}{x^2})}$$

To simplify this last fraction, multiply the numerator and denominator by $-x^4$:

$$= \lim_{x \rightarrow \infty} \frac{2x}{x^2 + 1} = 0$$

Finally, your answer is:

$$\exp\left[\lim_{n \rightarrow \infty} n\ln(1 + \frac{1}{n^2})\right] = \exp[0] = 1$$

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In the first two identities there should be noted that $\exp$ is a continuous function and thus we can pass the limit to the exponent. – Mussé Redi Apr 2 '14 at 18:13

It's really not necessary to know anything about the subtler limit, $e=\lim_{n\to\infty}\left(1+{1\over n}\right)^n$. A binomial expansion and some crude estimates are sufficient here. Note first that

$${n\choose k}\left({1\over n}\right)^k={n\over n}\cdot{n-1\over n}\cdot\cdots\cdot{1\over n}\le1$$

and therefore

$$\left(1+{1\over n^2}\right)^n=\sum_{k=0}^n{n\choose k}\left({1\over n^2}\right)^k=\sum_{k=0}^n{n\choose k}\left({1\over n}\right)^k{1\over n^k}\le\sum_{k=0}^n{1\over n^k}\le1+{1\over n}+{1\over n^2}(n-2)$$

The Squeeze Theorem takes over from here.

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