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I'm trying to demonstrate that $\left( 1+\frac1 n \right)^n$ is bigger than $2$. I have tried to prove that $\left( 1+\frac1 n \right)^n$ is smaller than $\left( 1+\frac1{n+1} \right)^{n+1}$ by expanding $\left( 1+\frac1n \right)^n = \sum\limits_{i=0}^n \left( \frac{n}{k} \right) \frac{1}{n^k}$ and $\left( 1+\frac1{n+1} \right)^{n+1} = \sum\limits_{i=0}^{n+1} \left( \frac{(n+1)}{k} \right) \frac{1}{(n+1)^k}$ but it doesn't seem to work.

What am I missing? Also, is there a method to demonstrate that without induction?

David G. Stork
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Sergi boquera fuentes
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  • Induction is going to be there, although it could be hiding in the proof of some other result. If you know the "expansion" (Binomial formula), then look at the first two terms $1+n\frac{1}{n}+...$. They already add up to $2$. Finally notice that the other terms are all non-negative. In this case induction is hiding in proving the binomial theorem. –  Jul 17 '18 at 00:08
  • In many books you can also find your inequality proven as a direct consequence of Bernoulli's inequality –  Jul 17 '18 at 00:10

4 Answers4

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You just need to consider the first $2$ terms in the binomial expansion \begin{eqnarray*} \left( 1+ \frac{1}{n} \right)^{n} = \sum_{i=0}^{n}\binom{n}{i}\frac{1}{n^i}=1+n \frac{1}{n} +\cdots \geq 2. \end{eqnarray*}

Donald Splutterwit
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Let $f(x) = (1+x)^n$. Note that $f$ is convex for $x \ge 0$ and so $f(x) \ge f(0)+f'(0) x = 1+xn$. Hence $f({1 \over n}) \ge 2$.

copper.hat
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\begin{align} \left(1+\frac1n\right)^n &= \sum_{k=0}^n {n \choose k} \frac1{n^k} \\ &= \sum_{k=0}^n \frac{n(n-1)\cdots(n-k+1)}{k!n^k} \\ &= \sum_{k=0}^n \frac1{k!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots \left(1-\frac{k-1}n\right) \end{align}

so

\begin{align} \left(1+\frac1{n+1}\right)^{n+1} &= \sum_{k=0}^{n+1} \frac1{k!}\left(1-\frac1{n+1}\right)\left(1-\frac2{n+1}\right)\cdots \left(1-\frac{k-1}{n+1}\right)\\ &> \sum_{k=0}^{n} \frac1{k!}\left(1-\frac1{n}\right)\left(1-\frac2{n}\right)\cdots \left(1-\frac{k-1}{n}\right)\\ &= \left(1+\frac1n\right)^n \end{align}

On the other hand, for $n = 1$ we have

$$\left(1+\frac11\right)^1 = 2$$

so $\left(1+\frac1n\right)^n > 2, \forall n \ge 2$.

mechanodroid
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Another way is to prove first that your sequence is monotonically increasing like has been done here:

I have to show $(1+\frac1n)^n$ is monotonically increasing sequence

... and since your first term is $2$, it follows that the subsequent ones are larger than $2$.

Momo
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