# Proving that $u_n = (1+ \frac{x}{n})^n$ is an increasing sequence for $n \in N$ and $n \geq |x|$.

I just want to check that my solution is correct.

I used Bernoulli's inequality to get the following result:

$$(1+ \frac{x}{n})^n \geq 1+n(\frac{x}{n}) = 1 + x$$ . Basically $$(1+ \frac{x}{n})^n \geq 1 + x$$.

Then I tried to show that the difference $$u_{n+1} - u_n \geq 0$$. So I did the following,

$$(1+ \frac{x}{n+1})^{n+1} - (1+ \frac{x}{n})^{n} \geq (1+ \frac{x}{n+1})^{n+1} - (1+x)$$ . (look two lines up)

But $$(1+ \frac{x}{n+1})^{n+1} \geq 1+(n+1)(\frac{x}{n+1}) = 1 + x$$ (the exact same thing as above).

So we conclude that $$u_{n+1} - u_n \geq 0$$.

Do you have any alternative simpler solutions? If so, it'd be great to share.

• How about looking at the derivative? Jan 12, 2022 at 8:49
• How did you deduce the second inequality? From $\left(1+\frac{x}{n}\right)^n\geq 1+x$, it does not follow that $-\left(1+\frac{x}{n}\right)^n\geq -(1+x)$. Jan 12, 2022 at 8:50
• Cf. this answer Jan 12, 2022 at 16:22

(1) Prove that $$\ln (1+u)\ge \frac{u}{u+1}$$ for $$u\ge 0$$.
(2) Replace $$u=\frac{x}{y}$$ in (1) and conclude that $$f(y)=(1+\frac{x}{y})^y$$ is increasing for $$y\ge x$$.
(3) Replace $$y=n$$ and finish the proof.