Proove that the sequence $\left (1+\frac{1}{x} \right )^{x}$ increases.

Question

I want to show that $\left (1+\frac{1}{x} \right )^{x}$ increases.

I have to show that $\left (1+\frac{1}{x+1} \right )^{x+1} > \left (1+\frac{1}{x} \right )^{x}$

$\left (1+\frac{1}{x}-\frac{1}{x(x+1)} \right )^{x+1} > \left (1+\frac{1}{x} \right )^{x}$

I triend to define $ \left (1+\frac{1}{x} \right ) = k$

$\left (k-\frac{1}{x(x+1)} \right )^{x+1} > k^x$

Now im pretty stuck.

for $n_1,n_2,\cdots,n_k$ use arithmatic and geometric means $$\frac{n_1+\cdots+n_k}{k}\geq\sqrt[k]{n_1\cdots n_k}$$ — Nosrati, Jan 16 '16 at 13:06
http://math.stackexchange.com/questions/83035/how-to-prove-11-xx-is-increasing-when-x0 http://math.stackexchange.com/questions/297916/proof-that-11-xx-is-monotonic-increasing — lab bhattacharjee, Jan 16 '16 at 13:06
@Noam If $x$ is in the real numbers, then (1) $f(x)$ isn't a sequence, and (2) knowing that $f(x+1)>f(x)$ is not enough to know that it is increasing. — Thomas Andrews, Jan 16 '16 at 13:51

AnatoliySultanov · Answer 1 · 2016-01-16T13:20:36.920

2

Let $x_n = (1 + \frac{1}{n})^n$. $\frac{x_{n+1}}{x_n} = \frac{n+2}{n+1}(\frac{1 + \frac{1}{n+1}}{1 + \frac{1}{n}})^n = \frac{n+2}{n+1} (\frac{n(n+2)}{(n+1)^2})^n = \frac{n+2}{n+1} (1 - \frac{1}{(n+1)^2})^n > \frac{n+2}{n+1}(1 - \frac{n}{(n+1)^2})$ (Bernoulli's inequality).

Then $\frac{n+2}{n+1}(1 - \frac{n}{(n+1)^2}) = 1 + \frac{1}{n+1} - \frac{n}{(n+1)^2} - \frac{n}{(n+1)^3} = 1 + \frac{1}{(n+1)^3} > 1$.

edited Jan 16 '16 at 13:20

answered Jan 16 '16 at 13:05

AnatoliySultanov

563

Proove that the sequence $\left (1+\frac{1}{x} \right )^{x}$ increases.

1 Answers1