Prove $2 < (1 + \frac{1}{n})^n$ for all $n \ge 2$

Question

Prove $2 < (1 + \frac{1}{n})^n$ for all $n \ge 2$. I believe this will come out by induction. It looks so much like induction, but I can't seem to find enough algebra to solve it. The base case: for $n=2$, we get $(1 + \frac{1}{2})^2 = 2 + \frac{1}{4} > 2$. Suppose now there is a $k \ge 2$ such that $(1 + \frac{1}{k})^k > 2$. I must show $(1 + \frac{1}{k+1})^{k+1} > 2$. From this, I tried to undo the exponent as in

\begin{align*} \left(1 + \frac{1}{k+1}\right)^{k+1} &= \left(1 + \frac{1}{k+1}\right)^k \left(1 + \frac{1}{k+1}\right) \end{align*}

Now I'm looking for a way to use $(1 + \frac{1}{k})^k > 2$, which is true by hypothesis. But it doesn't seem obvious how to turn a fraction $\frac{1}{k+1}$ into $\frac{1}{k}$.

By the way, I know $(1 + \frac{1}{n})^n$ approaches $e$ as $n$ approaches infinity, but I'm trying not to use any related identities to solve this problem. Can we do it by induction? What is the algebra that I'm missing there? Any hints? Thank you!

Thanks. Lots of good answers down there. I selected what seemed to be the most straightforward one. Though it looked to me like induction, it seems easier to apply the binomial theorem as @jjagmath kindly pointed out.

Solution. Considering just the first two terms of the binomial theorem expansion, we get

\begin{align*} (1 + 1/n)^n &= 1 + n (1/n) + ... + C(n,n) 1/n^n\\ &= 2 + ... + C(n,n) 1/n^n\\ &> 2. \end{align*}

We must restrict $n \ge 2$ because $n=1$ fails with $3/2 > 2$.

Answer 1

Hint: It's probably easier to prove the more general fact that $(1+x)^n>1+nx$ for all $n \ge 2$ and all $x>0$. You could do this by induction, or by using the binomial theorem.

Answer 2

Just use the Newton binomial theorem: $(1+\frac{1}{n})^n = 1 + \binom{n}{1}\frac{1}{n} + \cdots > 2$

Answer 3

$$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+...$$ and so: $$\left(1+\frac1n\right)^n=1+1+\frac1{2!}\frac{n-1}{n}+...$$ and so you can see that if $n=1$ then the value is $2$ and for $n\ge2$ there are non-zero terms after and so it has to be greater than two

Answer 4

Using Bernoulli's Inequality we can observe that $(1+x)^n>1+nx \implies$ $(1+\frac{1}{n})^n>1+\frac{1}{n}\cdot n=1+1=2$

Answer 5

Yet another method, using the AM-GM inequality:

First, write $1+\frac{1}{n}=\frac{n+1}{n}=\frac{2+\sum_{i=1}^{n-1} 1 }{n}$
Now we get by the AM-GM inequality:

$$ 1+\frac{1}{n}=\frac{2+\sum_{i=1}^{n-1} 1}{n}\geq \left(2\cdot\prod_{i=1}^{n-1} 1\right)^{\frac{1}{n}}=2^{\frac{1}{n}} $$

Since $2\neq 1$ the above inequality is strict for $n\geq 2$ (since for such $n$ we have both $2$ and $1$ in the means), now taking the $n$-th power of both sides gives the desired inequality (since $x\mapsto x^n$ is monotonically increasing):

$$ \left(1+\frac{1}{n}\right)^{n} > 2 $$