How do I prove this:

If Sn = (1+1/n)^n, then (sn) is strictly increasing.

I know the defn to be strictly increasing is: A sequence is strictly increasing if and only if for all n, Sn

Question

How do I prove this:

If Sn = (1+1/n)^n, then (sn) is strictly increasing.

I know the defn to be strictly increasing is: A sequence is strictly increasing if and only if for all n, Sn < Sn+1

anonymous · Answer

A proof by induction is one way to go about it.

First show it holds for some base case, say $n=1$:
$$s_n=\left(1+\frac{1}{n}\right)^n\
s_1=2\
s_2=\frac{9}{4}=2.25$$
$s_1<s_2$, so the base case is true.

Now assume the claim holds for $n=k$, i.e. that $s_k<s_{k+1}$. You must now show that this implies $s_{k+1}<s_{k+2}$.

Under the assumption, you have
$$\left(1+\frac{1}{k}\right)^k<\left(1+\frac{1}{k+1}\right)^{k+1}$$
You know that
$$s_{k+2}=\left(1+\frac{1}{k+2}\right)^{k+2}=\left(1+\frac{1}{k+2}\right)\left(1+\frac{1}{k+2}\right)^{k+1}$$
Now what you want to do is compare this with $s_{k+1}$. Here is what you need to establish:
$$\underbrace{\color{red}{\left(1+\frac{1}{k+1}\right)^{k+1}}}_{\large s_{k+1}}<\underbrace{\left(1+\frac{1}{k+2}\right)\color{red}{\left(1+\frac{1}{k+2}\right)^{k+1}}}_{\large s_{k+2}}$$
First compare the red quantities. Since they're raised to the same power, you can focus on what's inside the parentheses. As $k$ gets bigger, $1+\dfrac{1}{k+1}\approx1+\dfrac{1}{k+2}$. On the right side, the $1+\dfrac{1}{k+2}$ factor is greater than 1, which will constantly increase the value of the right side. This confirms that the right side will always be greater, thus completing the proof.