# Finding the limit of the sequence $x(n) = (1+2/n)^n$ [duplicate]

What we are allowed to use - 1) The fact that limit of $(1+1/n)^n$ exists and assumed to be some real number $e$ 2) Subsequencial properties of limits of sequences 3) Basic properties of limit

In the previous question $x(n) = (1+1/n^2)^{(2(n^2))}$

We just used the fact that the given sequence is square of the subsequence of the sequence $(1+1/n)^n$ And shall thus converge to $e^2$

I'd expect we'd be required to use something similar in this question but I am unable to, can't see a valid subsequence forming.

Edit - This question is not a duplicate of the question suggested as that question assumes that a lot more theorems have been proven, especially, Limits of log.

## marked as duplicate by user99914, Jimmy R., Claude Leibovici, N. F. Taussig, WatsonFeb 18 '16 at 12:35

• Try to rewrite you expression in the new variable $u=n/2$. – matovitch Feb 18 '16 at 12:00
• write $(1 + 2/n) = (n + 2)/(n + 1) \times (n + 1)/n$ and then on raising each factor to power $n$ we can see that each factor tends to $e$ and hence the overall product tends to $e^{2}$. – Paramanand Singh Feb 19 '16 at 4:08

The limit that defines the constant $e$ is $$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n$$ Since $x^2$ is a continuous function, we get $$e^2=\lim_{n\to\infty}\left(1+\frac2n+\frac1{n^2}\right)^n$$ It is simple to see that $$\left(1+\frac2n\right)^n\le\left(1+\frac2n+\frac1{n^2}\right)^n\le\left(1+\frac2n\right)^n\left(1+\frac1{n^2}\right)^n$$ which is the same as $$\left(1+\frac2n+\frac1{n^2}\right)^n\left(1-\frac1{n^2+1}\right)^n\le\left(1+\frac2n\right)^n\le\left(1+\frac2n+\frac1{n^2}\right)^n$$ which implies by Bernoulli's Inequality $$\left(1+\frac2n+\frac1{n^2}\right)^n\left(1-\frac{n}{n^2+1}\right)\le\left(1+\frac2n\right)^n\le\left(1+\frac2n+\frac1{n^2}\right)^n$$ The Squeeze Theorem says $$\lim_{n\to\infty}\left(1+\frac2n\right)^n=e^2$$

Let $n=2m$. Then $\lim_{n\to+\infty}x(n)=\lim_{m\to+\infty}x(2m)$ and $$\left(1+\frac{2}{n}\right)^n=\left(1+\frac{2}{2m}\right)^{2m}=\left(\left(1+\frac{1}{m}\right)^{m}\right)^2=\left(1+\frac{1}{m}\right)^{m}\left(1+\frac{1}{m}\right)^{m}$$

• How can we just assume that n=2m, n is not necessarily even. – Lelouch Feb 18 '16 at 12:25
• $m$ is not necessarily an integer. As we are taking the limit towards infinity, it does not matter. – nippon Feb 18 '16 at 12:38

Considering $\displaystyle \sqrt{x(2n)}=\left( 1+\frac{1}{n} \right)^{n}$,

as we know $\displaystyle \lim_{n\to \infty} \left( 1+\frac{1}{n} \right)^{n}=e$,

so $\displaystyle \lim_{n\to \infty} \sqrt{x(2n)}=e$,

$\displaystyle \lim_{n\to \infty} x(2n)=e^{2}$

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

• How is the first step valid? – Lelouch Feb 18 '16 at 12:27
• $\sqrt{x(2n)}=\sqrt{(1+\frac{2}{2n})^{2n}}=(1+\frac{1}{n})^{n}$ – Ng Chung Tak Feb 18 '16 at 12:32

Consider $$x_n=\Big(1+\frac 2n\Big)^n$$ Take logarithms $$\log(x_n)=n \log\Big(1+\frac 2n\Big)$$ Now, using that, for small $y$, $\log(1+y)\simeq y$, then $$\log(x_n)\simeq n \times \frac 2n=2$$

• Could you prove this by sequencial properties? Limits of log hasn't been proven yet. – Lelouch Feb 18 '16 at 12:26