How to proof (using by mathematical induction)($n\in \mathbb{N}$)

Question

I would appreciate it if somebody could help me with the following problem:

Q: How to proof (using by mathematical induction)($n=2,3,4,\cdots$)

$$\left(1+\frac{1}{n}\right)^n<1+\frac{1}{1!}+\frac{1}{2!}+\cdots+ \frac{1}{n!}$$

Answer 1

We invoke strong induction:

$\textbf{I:}$ We wish to show the statement is true for $n=2$. This can be done with a simple calculation of this inequality: $\left(1 + \frac{1}{2}\right)^2 < \left(1 + \frac{1}{1!} + \frac{1}{2!}\right)$.

$\textbf{II:}$ Suppose the above statement is true for $n \in [2,k-1]$, we wish to show the statement is true for $n=k$. From out hypothesis we know that $\left(1+\frac{1}{k-1}\right)^{k-1} < \left(1 + \frac{1}{1!} + \cdots + \frac{1}{(1-k)!}\right)$. Multiplying both sides by $1+\frac{1}{k-1}$, we have the string of inequalities $$ \left(1+\frac{1}{k}\right)^k < \left(1+\frac{1}{k-1}\right)^{k} < \left(1+\frac{1}{1!} + \cdots + \frac{1}{(k-1)!}\right) \times \left(1+\frac{1}{k-1}\right). $$ We can expand the right side of this inequality to obtain $$ 1+\frac{1}{1!}+\cdots+\frac{1}{(k-1)!} + \frac{1}{k-1} + \cdots + \frac{1}{(k-1)!(k-1)} = \sum_{i=1}^{k-1}\frac{k}{i!(k-1)}. $$ Finally, we can utilize the usual convergence tests to show that $$ \sum_{i=1}^{k-1}\frac{k}{i!(k-1)} < 1+ \sum_{1}^{k-1} \frac{1}{i!} + \frac{1}{k!}. $$

$\textbf{III:}$ Thus by the principle of strong induction, $\left(1+\frac{1}{n}\right)^n < \left(1 + \frac{1}{1!} + \cdots + \frac{1}{n!}\right)$ for all $n = 2, 3, \ldots \hspace{25pt} \square$