Using the binomial expansion, it is quite is easy to show that $$\left(1+\frac{1}{n}\right)^n \le \sum_{r=0}^{n} \frac{1}{r!} $$ for all $n\in\mathbb{Z^+}$, with equality holds when $n=1.$ (Can it be proved by the mathematical induction?)
But it seemes to me really difficult to prove that $$\sum_{r=0}^{n} \frac{1}{r!}\lt \left(1+\frac{1}{n}\right)^{n+1}$$ for all $n\in\mathbb{Z^+}$. Can anyone prove it?
As it can be proved that the sequences $\{(1+\frac{1}{n})^n\}$ and $\{(1+\frac{1}{n})^{n+1}\}$ converge to the same limit, the above inequalities will help establish the equivalence between the following definitions of e :
$$ \begin{align} e &= \lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n\\ e &= \sum_{r=0}^{\infty} \frac{1}{r!} \end{align} $$
2 Answers
Just $$\sum_{r=0}^{n} \frac{1}{r!}<\sum_{r=0}^{\infty} \frac{1}{r!}=e\lt \left(1+\frac{1}{n}\right)^{n+1}$$ because easy to show that $f(x)=\left(1+\frac{1}{x}\right)^{x+1}$ is a decreasing function on $(0,+\infty)$.
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4$\begingroup$ That rather defeats the purpose of the question $\endgroup$ Commented Jan 4, 2017 at 7:39
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$\begingroup$ My objective is to show that the two definitions of e are equivalent by proving the inequality first. So, the definitions of e are not allowed in the proof. $\endgroup$ Commented Jan 4, 2017 at 18:41
Answer Attempt: Inequality too Weak
Too long for comment, but I wanted to see if anyone could fix any errors in this proof. Feel free to make it community wiki if possible.
We assume that
$$\left(1+\frac{1}{n}\right)^{n} \le \sum_{r=0}^{n} \frac{1}{r!} \tag{1}$$
And we want to prove that
$$\left(\frac{n+2}{n+1}\right)^{n+2} \le \left(1+\frac{1}{n+1}\right)\sum_{r=0}^{n+1} \frac{1}{r!} \tag{2}$$
As usual, we assume this to be true. OP already proved $n=1$ is a base case. We now have that
$$\left(1+\frac{1}{n+1}\right)\sum_{r=0}^{n+1}\frac{1}{r!}$$
$$= \sum_{r=0}^{n+1}\frac{1}{r!}+\frac{1}{n+1}\sum_{r=0}^{n+1}\frac{1}{r!}$$
$$= \frac{1}{(n+1)!}+\sum_{r=0}^{n}\frac{1}{r!}+\frac{1}{(n+1)(n+1)!}+\frac{1}{n+1}\sum_{r=0}^{n}\frac{1}{r!}$$
$$\ge \frac{n+2}{(n+1)^2 n!}+\left(1+\frac{1}{n}\right)^n\left(\frac{n+2}{n+1}\right)$$ Unfortunately, it looks like the inequality bounds are barely not tight enough, because this final term is smaller than we need it to be. Given that I appealed only to the induction hypothesis so far this is midly surprising; perhaps the argument can be fixed from here?
