Computing $\lim\limits_{n\to\infty}(1+1/n^2)^n$

Question

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

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

Answer 1

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

Answer 2

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 $e^{1/n}$.

Answer 3

Hint: prove that $$ \log (1+u) \le u $$

then use the continuity of $\exp$.

details:

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

Answer 4

You can do this without knowing anything about $e$. Using Bernoulli's Inequality we have $$\left(1 - \frac{1}{n^{4}}\right)^{n} \geq 1 - \frac{1}{n^{3}}$$ and clearly we have $$\left(1 - \frac{1}{n^{4}}\right)^{n} \leq 1 - \frac{1}{n^{4}}$$ so that $$1 - \frac{1}{n^{3}} \leq \left(1 - \frac{1}{n^{4}}\right)^{n} \leq 1 - \frac{1}{n^{4}}\tag{1}$$ and via Squeeze theorem we get $$\lim_{n \to \infty}\left(1 - \frac{1}{n^{4}}\right)^{n} = 1\tag{2}$$ Using similar technique we get the inequality $$1 - \frac{1}{n} \leq \left(1 - \frac{1}{n^{2}}\right)^{n} \leq 1 - \frac{1}{n^{2}}$$ and via Sqeeze theorem we get $$\lim_{n \to \infty}\left(1 - \frac{1}{n^{2}}\right)^{n} = 1\tag{3}$$ and dividing $(2)$ by $(3)$ we get $$\lim_{n \to \infty}\left(1 + \frac{1}{n^{2}}\right)^{n} = 1\tag{4}$$

Answer 5

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.

Answer 6

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:

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