$(1+\frac{1}{n})^n \le 1+1+ \frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}$

Question

My book is introducing sequences and series and presents this, for $n \ge 1$ $$\begin{align*} \left(1+\frac{1}{n}\right)^n&=1+\binom{n}{1}\frac{1}{n}+\binom{n}{2}\frac{1}{n^2}+\binom{n}{3}\frac{1}{n^3}+\dots+\binom{n}{n}\frac{1}{n^n}\\ &=1+1+\frac{n(n-1)}{2}\cdot\frac{1}{2!}+\frac{n(n-1)(n-2)}{n^3}\cdot\frac{1}{3!}+\dots+\frac{n^n}{n!}\cdot\frac{1}{n!} \end{align*}$$

Then it states that this implies $$\left(1+\frac{1}{n}\right)^n\leq1+1+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{n!}$$

However, I can't see how to get this inequality from the first part. Thanks for any help.

Answer 1

The missing bit should be given by this inequality: $$\begin{align*} \binom{n}{k}\frac{1}{n^k}&=\frac{n\cdot(n-1)\dots(n-k+1)}{k!}\frac{1}{n^k}\\ &\leq \frac{n^k}{k!}\frac{1}{n^k} = \frac{1}{k!} \end{align*}$$ where the equality is explained by "enlarging" every factor to $n$.

Answer 2

Let label the right side of the first equation as ${R_1}$ and that of the second equation ${R_2}$. Also, let the left side be labeled as $L$.

Each term on the right in the first equation is equal to or smaller than the corresponding term in the second equation.

Here's an example showing how. Consider the third term.

$$\begin{align*} \frac{n(n-1)}{n^2} = \frac{n-1}{n} < 1 \end{align*}$$

That means,

$$\begin{align*} \frac{n(n-1)}{n^2} \cdot \frac{1}{2!} = \frac{n-1}{n} \cdot \frac{1}{2!} < \frac{1}{2!} \end{align*}$$

[Observe the fraction containing $n$. The numerator ($n-1$) is smaller then the denominator ($n$).]

So, $R_1 \le R_2$. And since $L = R_1$, $L \le R_2$.

Answer 3

In cases like this, it is helpful to compare term by term until you notice a pattern. For example, we have $$(1+\frac{1}{n})^{n} = \sum_{i=0}^k \binom{n}{i} \frac{1}{n^i} \leq \sum_{i=0}^k\frac{1}{i!}$$ Although this may seem daunting at first, we can compare $\binom{n}{i} \frac{1}{n^i}$ and $\frac{1}{i!}$ directly. Note that your textbook showed that $$\binom{n}{i} \frac{1}{n^i} = \frac{n\cdot(n-1)\cdot\dotsc\cdot(n-i+1)}{n^i} \frac{1}{i!}$$ So ultimately you are comparing the term above and $\frac{1}{i!}$. Hopefully you might be able to convince yourself that if $n \geq 1$, then $\frac{n\cdot(n-1)\cdot\dotsc\cdot(n-i+1)}{n^i} \leq 1$ since you are essentially multiplying numbers fractions whose values are between $0$ and $1$. So, to make things easier let's make a variable $c_{i}$ such that $c_{i} = \frac{n\cdot(n-1)\cdot\dotsc\cdot(n-i+1)}{n^i}$. Now we can rewrite our sum as $$(1+\frac{1}{n})^{n} = \sum_{i=0}^k \frac{c_i}{i!} \leq \sum_{i=0}^k\frac{1}{i!}$$ Since we know $c_i \leq 1$ for every term, the inequality holds true.