e.md

We want to prove that

$$ \left(1+\dfrac{1}{n}\right)^n < e < \left(1+\dfrac{1}{n}\right)^{n+1} $$

We will assume that $n \in \mathbb{N}$, $n &gt; 0$.

Left Side $(1+\dfrac{1}{n})^n &lt; e$

We already know that $e = \text{lim}_{n \rightarrow \infty} (1 + \dfrac{1}{n})^n$. In that limit, $(1 + \dfrac{1}{n})^n$ is effectively the left side of the above inequality.

This implies that $(1+\dfrac{1}{n})^n$ has an asymptote at $e$, as $n\rightarrow\infty$.

The only way that the inequality will hold is if $(1+\dfrac{1}{n})^n$ increases monotonically. Put another way, for all $n$, there exists a $n + 1$ such that $(1+\dfrac{1}{n+1})^{n+1} &gt; (1+\dfrac{1}{n})^{n}$. If that's the case, then we know that for any $n$, $(1+\dfrac{1}{n})^n &lt; e$

To prove this, let's use the Bernoulli's Inequality.

$$ (1 + x)^n \ge 1 + nx $$

We shall prove by induction that $(1 + \dfrac{1}{n})^n &gt; (1 + \dfrac{1}{n - 1})^{n-1}$, can be algebraically manipulated to yield the Bernoulli's Inequality.

$$ \left(1 + \dfrac{1}{n}\right)^n > \left(1 + \dfrac{1}{n - 1}\right)^{n-1} \Rightarrow $$

$$ \left(1 + \dfrac{1}{n}\right)^n\left(1 + \dfrac{1}{n - 1}\right)^{-n} > \left(1 + \dfrac{1}{n - 1}\right)^{-1} \Rightarrow $$

$$ \left(\left(1 + \dfrac{1}{n}\right)\left(1 + \dfrac{1}{n - 1}\right)^{-1}\right)^n > \left(1 + \dfrac{1}{n - 1}\right)^{-1} \Rightarrow $$

$$ \left(\dfrac{n+1}{n}\cdot\left(\dfrac{n}{n-1}\right)^{-1}\right)^n > \left(\dfrac{n}{n - 1}\right)^{-1} \Rightarrow $$

$$ \left(\dfrac{n+1}{n}\cdot\dfrac{n-1}{n}\right)^n > \dfrac{n-1}{n} \Rightarrow $$

$$ \left(\dfrac{n^2-1}{n^2}\right)^n > \dfrac{n-1}{n} $$

If we are to look at the left hand side, it is effectively

$$ \left(\dfrac{n^2-1}{n^2}\right)^n = \left(1-\dfrac{1}{n^2}\right)^n $$

Likewise for right-hand side

$$ \dfrac{n-1}{n} = 1 - \dfrac{1}{n} $$

Putting it all together, the inequality finally looks like so

$$ \left(1-\dfrac{1}{n^2}\right)^n > 1 - \dfrac{1}{n} $$

By Bernoulli's Inequality, let $x = -\dfrac{1}{n^2}$, then the left hand side is $(1 + x)^n$, and, the right-hand side is $1 + nx$, thus giving us the inequality

$$ (1 + x)^n > 1 + nx \Rightarrow $$

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

Where $1 + n(-\dfrac{1}{n^2}) = 1-\dfrac{1}{n}$.

Thus showing what needs to be shown.

Now, because we've proved that the left side is monotonically increasing, then for all values $n$, $(1+\dfrac{1}{n})^n &lt; e$.

Right Side $e &lt; (1+\dfrac{1}{n})^{n+1}$

We know that $\text{lim}_{n\rightarrow\infty}(1+\dfrac{1}{n})^{n+1} = e$.

Proof:

$$ \lim_{n\rightarrow\infty}\left(1+\dfrac{1}{n}\right)^{n+1} = \lim_{n\rightarrow\infty}\left(1+\dfrac{1}{n}\right)^n\left(1+\dfrac{1}{n}\right) = \lim_{n\rightarrow\infty}\left(1+\dfrac{1}{n}\right)^n\lim_{n\rightarrow\infty}\left(1+\dfrac{1}{n}\right) = \lim_{n\rightarrow\infty}\left(1+\dfrac{1}{n}\right)^n\left(\lim_{n\rightarrow\infty}1+\lim_{n\rightarrow\infty}\dfrac{1}{n}\right) $$

This implies that $(1+\dfrac{1}{n})^{n+1}$ has an asymptote at $e$, as $n\rightarrow\infty$.

It is suffice to say that

$$ \left(1+\dfrac{1}{n}\right)^n < \left(1+\dfrac{1}{n}\right)^{n+1} \Rightarrow $$

$$ \dfrac{\left(1+\dfrac{1}{n}\right)^n}{\left(1+\dfrac{1}{n}\right)^n} < \dfrac{\left(1+\dfrac{1}{n}\right)^{n+1}}{\left(1+\dfrac{1}{n}\right)^n} \Rightarrow $$

$$ 1 < 1+\dfrac{1}{n} $$

But some may be skeptical of that proof.

We can take the log of all parts.

$$ \ln\left(1 + \dfrac{1}{n}\right)^n < \ln e < \ln\left(1 + \dfrac{1}{n}\right)^{n+1} \Rightarrow $$

$$ n\ln\left(1 + \dfrac{1}{n}\right) < 1 < (n+1)\ln\left(1 + \dfrac{1}{n}\right) \Rightarrow $$

$$ n < \dfrac{1}{\ln\left(1 + \dfrac{1}{n}\right)} < n+1 $$

But that still isn't convincing.

So, the best way to prove the inequality is to also prove that $(1+\dfrac{1}{n})^{n+1}$ is monotonically decreasing.

We will use the same Bernoulli's Inequality to prove the inequality.

We want to show that for all $n$, there exists an $n+1$ such that

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

Manipulating the above algebraically, we get

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

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

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

$$ \left(\dfrac{n+1}{n}\right)^{-1}\left(\dfrac{n+1}{n+2}\cdot\dfrac{n+1}{n}\right)^{n+2} > 1 \Rightarrow $$

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

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

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

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

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

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

Same procedure, again. We want to re-interpret the above resulting inequality to Bernoulli's Inequality, of the form $(1 + x)^n &gt; 1 + xn$.

We want to reinterpret $x$ as $\dfrac{1}{n(n+2)}$, and $n$ as $n + 2$. The left hand side is trivially equivalent. For the right hand side where $1 + xn$ is congruent with $1 + \dfrac{1}{n(n+2)}(n+2)$.

Trivially

$1 + \dfrac{1}{n(n+2)}(n+2) = 1 + \dfrac{1}{n}$

Thus proving that

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

Or, in words, that $\left(1+\dfrac{1}{n}\right)^{n+1}$ is monotonically decreasing for all $n$, and $n+1$.

Now because $\left(1+\dfrac{1}{n}\right)^{n+1}$ has an asymptote at $e$ as $n \rightarrow \infty$, and that it is monotonically decreasing, then for all $n$, $\left(1+\dfrac{1}{n}\right)^{n+1} &gt; e$.

Sources

[0] https://math.stackexchange.com/questions/167843/show-that-left1-dfrac1n-rightn-is-monotonically-increasing
[1] https://math.stackexchange.com/questions/2071492/the-function-f-n-1-1-n-n1-is-decreasing
[3] https://math.stackexchange.com/questions/306178/given-y-n-1-frac1nn1-show-that-lbrace-y-n-rbrace-is-a-decreasin