Prove the inequality $\left(1+\frac{1}{n}\right)^n < \left(1+\frac{1}{n+1}\right)^{n+1}$

Question

Can anybody show me the way of proving it without induction.My textbook showed the way but it was unclear for me.It used the statements such as $(n+1)a^n < b^n +b^{n-1}a.....+a^n < (n+1)b^n$ I don't know how they come to this statement I know it's a duplicate question but the type of answer I am searching for should be derived from above statement so that I could understand how author intended to prove

Answer 1

$(1+\frac 1n)^n = 1 + \frac {n}{n} + \frac {n(n-1)}{2n^2} + \frac {n(n-1)(n-2)}{3!n^3} + \cdots + \frac {n!}{n!n^n}$ by the binomaly theorem

And with a little bit of simplification we get:

$1 + 1 + \frac 12(1-\frac{1}{n}) + \frac 1{3!}(1-\frac{1}{n})(1-\frac{2}{n})+\cdots+\frac{1}{n!}(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{n-1}{n})$

While

$(1+\frac 1{n+1})^{n+1} = 1 + 1 + \frac 12(1-\frac{1}{n+1}) + \frac 1{3!}(1-\frac{1}{n+1})(1-\frac{2}{n+1})+\cdots+\frac{1}{n!}(1-\frac{1}{n+1})(1-\frac{2}{n+1})\cdots(1-\frac{n-2}{n+1}) + \frac{1}{(n+1)!}(1-\frac{1}{n+1})(1-\frac{2}{n+1})\cdots(1-\frac{n-1}{n+1})$

Every term of both expansions is strictly positive.

Comparing term by term each term of $(1+\frac{1}{n+1})^{n+1}$ is greater than or equal to the same term of $(1+\frac1{n})^{n}.$ And $(1+\frac{1}{n+1})^{n+1}$ has one additional positive term.

Answer 2

Differentiating $(1+\frac{1}{x})^{x}$ gives $$\frac{d}{dx}\left(1+\frac{1}{x}\right)^{x} = \left(1+\frac{1}{x}\right)^{x}\left(\ln\left(1+\frac{1}{x}\right) -\frac{1}{x+1}\right)$$

Since $\left(1+\frac{1}{x}\right)^{x} > 0$ for all real $x \not= 0$ and $\left(\ln\left(1+\frac{1}{x}\right) -\frac{1}{x+1}\right) > 0$ for all real $x \not= 0$, the function is continuously increasing from $(-\infty, 0) \cup (0, \infty)$.

Therefore, $$\left(1+\frac{1}{n}\right)^{n} < \left(1+\frac{1}{n+1}\right)^{n+1}$$ for positive $n$.