1

I'm stuck with this proof, this is what I did:

$n=2: \bigg(1+\cfrac{1}{2}\bigg)^{2} < 1 + 1 + \cfrac{1}{2}$

where verifies: $\cfrac{9}{4} < \cfrac{5}{2}$

Now with the hypothesis: $\bigg(1+\cfrac{1}{k}\bigg)^{k} < \sum_{i=0}^k \cfrac{1}{i!}$

then: $\bigg(1+\cfrac{1}{k+1}\bigg)^{k+1} < \sum_{i=0}^{k+1} \cfrac{1}{i!}$

$\bigg(1+\cfrac{1}{k+1}\bigg)^{k+1}=\bigg(1+\cfrac{1}{k+1}\bigg)^{k}\bigg(1+\cfrac{1}{k+1}\bigg)<\bigg(1+\cfrac{1}{k}\bigg)^{k}\bigg(1+\cfrac{1}{k+1}\bigg)<\sum_{i=0}^{k} \cfrac{1}{i!}\bigg(1+\cfrac{1}{k+1}\bigg)=\sum_{i=0}^{k} \cfrac{1}{i!} +\cfrac{1}{k+1}\sum_{i=0}^{k} \cfrac{1}{i!}$

and i don't know how to relate the second element and get $\cfrac{1}{(k+1)!}$. Should I use Newton binomial?

Thanks in advance.

Calvin Khor
  • 34,903
Shiro_P
  • 33
  • 1
    See https://math.stackexchange.com/q/404916/42969 – Martin R Jun 23 '23 at 03:42
  • 1
    In principle, using $$\left(1+\frac1{k+1}\right)^k < \left(1+\frac1k\right)^k$$ is not strong enough. If we just induct on that, we'd have the estimate $\left(1+\frac1{k+1}\right)^{k+1} < \frac{3(k+2)}4$, which isn't good enough. – Brian Moehring Jun 23 '23 at 03:42

1 Answers1

2

Although not by induction, here is a proof,

$$(1+1/n)^n = \sum_{i=0}^n {n \choose i} \frac{1}{n^i} = \sum_{i=0}^n \frac{1}{i!} \frac{n(n-1)(n-2)...(n-i+1)}{n^i} < \sum_{i=0}^n \frac{1}{i!} $$

Balaji sb
  • 4,357