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

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?

• $A^n=[A^{n^2}]^{1/n}$ – Piquito Dec 6 '15 at 14:29
• we can write $(1 + \frac{1}{n})^n \to e = 2.17$ but if that fraction is very small we just get $1$. As you are seeing $(1 + \frac{1}{n^2})^n \to 1$. It's that $\frac{1}{n^2}$ is just way too small. – cactus314 Dec 6 '15 at 19:39

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

• While your answer is correct, it lacks proper justification because asymptotically equivalent expressions cannot be just substituted anywhere. – user21820 Dec 6 '15 at 14:01
• Yes, it's a hint for the OP. – Olivier Oloa Dec 6 '15 at 14:20
• Well I don't recommend doing that because that's how so many students do all kinds of incorrect manipulation because they think "Looks similar". To be more specific, the error terms that Gudson provided for the log are crucial in establishing that they don't affect the result. – user21820 Dec 6 '15 at 14:24
• Thank you very much for your recommendation. – Olivier Oloa Dec 6 '15 at 14:27

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 $n \rightarrow \infty$.

• In fact there's an even easier way to see that $e^x \ge 1+x$.. Just check that they coincide at $x = 0$ and compare derivatives of both sides of the inequality. – user21820 Dec 6 '15 at 14:03

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 $n \to \infty$.

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 $n \to\infty$.

• This is much better than all the other arguments which involve $e$ – Trajan Dec 6 '15 at 15:27

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

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

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

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

• 5th inequality has no justification that would be obvious to the OP. (use of derivatives) – Trajan Dec 6 '15 at 15:26

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 $1$.

• 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.
• Is this too heuristic? I dont like the upper bound method – Trajan Dec 6 '15 at 15:21
• @Permian: Why would it be? It's a rigorous argument as far as I'm aware. – hmakholm left over Monica Dec 6 '15 at 15:22
• $\left(1+\frac1{n^2}\right)^{n^2} < 3$ has not been justified. Its not wrong per se, it just assumes niceness of the exponential function a priori to using it. Actually I do believe we need to know your bullet point method to make the answer "more right" – Trajan Dec 6 '15 at 15:29
• Its unlikely in an undergraduate course on real analysis you would use the argument that $e=2.718$.... – Trajan Dec 6 '15 at 15:44
• @Permian: Well, I agree that this answer is using too much, since we could have just used $e+1$ instead of $3$... – user21820 Dec 10 '15 at 0:15

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