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

Question

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 did you deduce the second inequality? From $\left(1+\frac{x}{n}\right)^n\geq 1+x$, it does not follow that $-\left(1+\frac{x}{n}\right)^n\geq -(1+x)$. — Alann Rosas, Jan 12 '22 at 08:50

score 1 · Answer 1 · answered Jan 12 '22 at 15:54

1

Plan for proof

(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.

answered Jan 12 '22 at 15:54

Mostafa Ayaz

1 Answers1