3

What's an easy way to find out the limit of $a_n = \left (1+\frac{1}{n^2} \right )^n$ for $n \rightarrow \infty$?

I don't think binomial expansion like with $\left (1-\frac{1}{n^2} \right )^n = \left (1+\frac{1}{n} \right )^n \cdot \left (1-\frac{1}{n} \right )^n$ is possible.

And Bernoulli's inequality only shows $\left (1+\frac{1}{n^2} \right )^n \geq 1 + n \cdot \frac{1}{n^2} = 1 + \frac{1}{n} \geq 1$ which doesn't seem to help as well.

Steph-P
  • 165

8 Answers8

4

You could have just committed to your original idea: $$\lim_{n\to\infty} \left(1+\frac{1}{n^2}\right)^n = \lim_{n\to\infty}\left(1+\frac{i}{n}\right)^n\cdot \lim_{n\to\infty}\left(1-\frac{i}{n}\right)^n = e^i \cdot e^{-i} = 1$$

adfriedman
  • 3,641
  • 3
    It's probably quite a task in and of itself to show that $e^z=\lim_{n\to\infty}\left(1+\frac zn\right)^n$ holds for complex $z$. – Simply Beautiful Art May 16 '17 at 23:31
  • @SimplyBeautifulArt It follows from the identity theorem if you accept it for real $z$ – user420261 May 16 '17 at 23:37
  • @user420261 I don't think invoking the identity theorem should really be necessary for a basic limit problem :P – Simply Beautiful Art May 16 '17 at 23:43
  • Equivalent Characterizations of exp.

    I'm fairly sure all characterizations are discussed in any elementary complex analysis class. $e^z$ normally behaves well enough for assumptions. I know maintaining some mysticism about swapping in and out of the complex domain in high school made analysis a lot more exciting for me to learn.

     – adfriedman May 16 '17 at 23:44
  • 1
    @adfriedman I doubt the OP has taken complex analysis courses, but you never know, I could be wrong :-) – Simply Beautiful Art May 16 '17 at 23:45
  • That isn't really the point. You can define $e^z = \lim_{n\to\infty}(1+z/n)^n$ for all $z$ and derive the power series and differential equation forms from there. I doubt the OP has seen the existence proof of $e$ by limit definition anyway, so nitpicking on what is "too advanced" is a bit silly. – adfriedman May 17 '17 at 00:07
  • There is a much easier way that circumvents appealing to the exponential function. – Mark Viola May 17 '17 at 00:14
  • @adfriedman How can you prove that $e^ze^{-z}=1$ then? That is, defining it to be the limit is only the first step of many things we need to subsequently prove. – Simply Beautiful Art May 17 '17 at 00:24
  • @SimplyBeautifulArt Hint: Try multiplying $\sum_{n=0}^{\infty} \frac{z^n}{n!}$ and $\sum_{n=0}^{\infty} \frac{(-z)^n}{n!}$. – adfriedman May 17 '17 at 01:51
  • @MarkViola Perhaps more elementary, but I disagree with "easier". For those willing to make the leap, it is trivial to follow. Indeed it looks like the OP considered this, but didn't make the leap to the complex plane. – adfriedman May 17 '17 at 02:00
  • I think using the equation $e^{z} = \lim_{n \to \infty}(1 + (z/n))^{n}$ for all $z\in\mathbb{C}$ is a fantastic idea with more intuitive appeal than defining it via power series. But this definition does require some effort to prove things. The relation $e^{z}e^{-z} = 1$ can be proved directly without going the series way and one can justify your answer on the basis of that, but I think the solution by Mark Viola is simpler (but not perhaps as obvious as yours) as it avoids any complicated stuff like $e$. – Paramanand Singh May 17 '17 at 10:30
4

There is an easy inductive proof that $(1+{1\over N})^N\le3-{1\over N}$ for all $N\in\mathbb{N}$, from which $(1+{1\over N})^N\lt3$ for all $N$ follows. Letting $N=n^2$, this tells us that

$$\left(1+{1\over n^2}\right)^n\lt3^{1/n}$$

If you now prove that $3^{1/n}\to1$ as $n\to\infty$, the Squeeze Theorem says that $(1+{1\over n^2})^n\to1$ as well.

Barry Cipra
  • 79,832
4

One of the most useful inequalities in asymptotic analysis is $$e^x \ge 1+x$$ (with near equality for small $x$, and equality for $x=0$). Here, it tells us that $1 + \frac1{n^2} \le e^{1/n^2}$, so $\left(1 + \frac1{n^2}\right)^n \le e^{1/n}$. As $n \to \infty$, $\frac1n \to 0$ and $e^{1/n} \to 1$; on the other hand, $\left(1+\frac1{n^2}\right)^n > 1$ for all $n$, so we're done by the squeeze theorem.

(In general, the inequality $e^x \ge 1+x$ helps us turn everything into a product while maintaining an upper bound that's fairly tight when $x$ is small.)

Misha Lavrov
  • 142,276
4

Here is a purely algebraic approach. First we note that

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

From $(1)$ it is easy to see that

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

Applying Bernoulli's Inequality to the term on the left-hand side of $(2)$ reveals

$$1\le \left(1+\frac1{n^2}\right)^n\le \frac1{1-\frac1n}$$

whereupon applying the sqeeze theorem yields the covetes limit

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

Mark Viola
  • 179,405
2

Using the binomial theorem and the fact that $\binom{n}{k}\leq\frac{n^k}{k!}$, we have: $$1\leq\left(1+\frac1{n^2}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac1{n^{2k}}\leq\sum_{k=0}^n\frac{n^k}{k!n^{2k}}=\sum_{k=0}^n\frac{1}{k!n^k}\leq1+\frac1n+\sum_{k=2}^n\frac1{n^2}\leq1+\frac2n.$$ Therefore, by the squeeze theorem, the limit is $1$.

Dejan Govc
  • 17,007
1

Use equivalents from the following equality: $$a_n=\exp\left(n\ln\left(1+\frac{1}{n^2}\right)\right).$$ Recall that $\ln(1+x)=x+o(x)$ and that $\exp$ is continuous.

C. Falcon
  • 19,042
1

Simple:

$$a_n=\left(\left(1+\frac1{n^2}\right)^{n^2}\right)^{1/n}\stackrel{n\to\infty}\longrightarrow e^{1/\infty}=e^0=1$$

Using the historically first definition of $e$:

$$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n=\lim_{n\to\infty}\left(1+\frac1{n^2}\right)^{n^2}$$

Simply Beautiful Art
  • 74,685
0

If $n \to \infty$ the this becomes form of $1^{\infty}$ to which you get, $e^{\lim_{n \to \infty} \frac{1}{n}}$ and you get $e^{0}$ which is 1.

Em.
  • 15,981
Iti Shree
  • 1,476