Prove that $( 1 + n^{-2}) ^n \to 1$.

Question

I need to prove that $\left(1 + \frac 1 {n^2} \right)^n \to 1$.

I tried to use Bernoulli's inequality, but that is not very useful since in the original sequence there is a plus sign. I then tried to use the Sandwich Theorem by finding two sequences which would make bounds for the original one. The lower bound is obvious, the upper bound not so much. I tried using the sequence $\left( \frac 1 {n+1} \right) ^{\frac 1 n}$, but I could not show that this sequence is bigger than the original one for all $n$. Could anyone help me with this?

Answer 1

One may write, as $n \to \infty$, $$ \left(1+\frac1{n^2} \right)^n=e^{\large n\log\left(1+\frac1{n^2} \right) }\sim e^{\large n\times\frac1{n^2} }=e^{\large \frac1{n}} \to 1. $$

Answer 2

Using Taylor approximation to $\log(1 + x)$ for $x \to 0$ gives $$ (1+n^{-2})^{n} = \exp [ n \log (1 + n^{-2})] = \exp [n\cdot n^{-2} + n\cdot n^{-2}\cdot o(1)] = \exp \bigg[ \frac{1}{n} + n^{-1}o(1) \bigg] \to e^{0} = 1 $$ as $n \to \infty$.

Answer 3

Note that $$(1+\frac{1}{n^2})^n -1 = \frac{1}{n^2}\sum_0^{n-1}(1+\frac{1}{n^2})^l \le \frac{1}{n^2} ne$$

Answer 4

For any $x > 0$:

  $\frac{1}{n^2} \le \frac{x}{n}$ as $n \to \infty$.

  Thus $(1+\frac{1}{n^2})^n \le (1+\frac{x}{n})^n \to e^x$ as $n \to \infty$.

  Thus $\limsup_{n\to\infty} (1+\frac{1}{n^2})^n \le e^x$.

Therefore $\limsup_{n\to\infty} (1+\frac{1}{n^2})^n \le 1$ because $e^x \to 1$ as $x \to 0^+$.

Clearly $\liminf_{n\to\infty} (1+\frac{1}{n^2})^n \ge 1$.

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

Answer 5

Use $e^x \geq 1 + x$ for all $x \in \mathbb{R}$. There are many ways to see this. The easiest, in my opinion, is to note that $y = x + 1$ is tangent to $y = e^x$ and $x \mapsto e^x$ is convex, then use definition of convexity.

It follows that $$ 1 \leq \left(1 + \frac{1}{n^2}\right)^n \leq \left(\exp\frac{1}{n^2}\right)^n = \exp\frac{1}{n} \rightarrow 1 $$ as $n \rightarrow \infty$.

Answer 6

We can do this using just the binomial expansion, without involving $e$. Note that if $r \ge 1$, then

$$\binom{n}{r} = \dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r(r-1)(r-1)\cdots 2\cdot 1} \le n(n-1)(n-2)\cdots(n-r+1)\le n^r$$

and it's true for $r=0$, too, because $1 \le 1$. So the binomial theorem gives

$$\left(1+\frac{1}{n^2}\right)^n = \sum_{r=0}^n\binom{n}{r}\frac{1}{n^{2r}} \le \sum_{r=0}^n \frac{1}{n^r} < \sum_{r=0}^\infty \frac{1}{n^r} = \frac{1}{1-1/n} $$

which tends to $1$ as $n \to\infty$.

Answer 7

$$\lim_{n\to\infty}\left(1+\frac{1}{n^2}\right)^n=$$ $$\lim_{n\to\infty}\exp\left(\ln\left(\left(1+\frac{1}{n^2}\right)^n\right)\right)=$$ $$\lim_{n\to\infty}\exp\left(n\ln\left(1+\frac{1}{n^2}\right)\right)=$$ $$\exp\left(\lim_{n\to\infty}n\ln\left(1+\frac{1}{n^2}\right)\right)=$$ $$\exp\left(\lim_{n\to\infty}\frac{\ln\left(1+\frac{1}{n^2}\right)}{\frac{1}{n}}\right)=$$ $$\exp\left(\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\left(\ln\left(1+\frac{1}{n^2}\right)\right)}{\frac{\text{d}}{\text{d}n}\left(\frac{1}{n}\right)}\right)=$$ $$\exp\left(\lim_{n\to\infty}\frac{-\frac{2}{n^3\left(1+\frac{1}{n^2}\right)}}{-\frac{1}{n^2}}\right)=$$ $$\exp\left(\lim_{n\to\infty}\frac{2n}{n^2+1}\right)=$$

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Since $2n$ grows asymptotically slower than $n^2+1$ as $n$ approaches $\infty$, $\lim_{n\to\infty}\frac{2n}{n^2+1}=0$:

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$$\exp\left(0\right)=e^0=1$$

Answer 8

Note that $$ a_n = \left(1+\frac1{n^2}\right)^n = \left(\left(1+\frac1{n^2}\right)^{n^2}\right)^{1/n} $$

The inner power on the right-hand side converges to $e$, so when $n$ is sufficiently large* we have $$ 1 < \left(1+\frac1{n^2}\right)^{n^2} < 3 $$ From this point on, therefore, we have $$ 1 < a_n < \sqrt[n]3 $$ This squeezes the original limit between two sequences that converge to $1$.

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• In fact every $n\ge 1$ is sufficiently large because $(1+1/n)^n$ approaches $e$ from below -- but we don't need to know that here.