Show that $x_n = \left(1+{x\over n}\right)^{n+k}$ is decreasing for $0<x<k$; $n, k\in \mathbb N$

Question

I've been working on some classical proofs of the sequences in the form $\left(1+{x\over n}\right)^p$.

So:

Let $n,k \in \mathbb N$ and: $$ \begin{cases} x_n = \left(1+{x\over n}\right)^{n+k}\\ 0 < x < k \end{cases} $$ Prove $\{x_n\}$ is a decreasing sequence.

$\Box$ One of the ways to see how the sequence behaves is to find a fraction of consecutive terms:

$$ \frac{x_n}{x_{n+1}} = \frac{\left(1+{x\over n}\right)^{n+k}}{\left(1+{x\over n + 1}\right)^{n+k+1}} = \left(\frac{n^2+nx+n+x}{n^2+nx+n}\right)^{n+k}\cdot \frac{1}{1+\frac{x}{n+1}} =\\ =\left(1+\frac{x}{n^2+nx+n}\right)^{n+k}\cdot \frac{1}{1+\frac{x}{n+1}} \stackrel{\text{Bernoulli's}}{>}\\ >\left(1+\frac{x(n+k)}{n(n+x+1)}\right)\cdot \frac{1}{1+\frac{x}{n+1}} $$ We know that $k>x$ therefore (note how $x$ changes to $k$ in the denominator): $$ \frac{x_n}{x_{n+1}} > \left(1+\frac{x(n+k)}{n(n+k+1)}\right)\cdot \frac{1}{1+\frac{x}{n+1}} $$

Here I wasn't able to notice some "smart" substitution and just expanded the terms in order to obtain:

$$ \frac{x_n}{x_{n+1}} > \frac{kn^2+knx+kn+n^3+xn^2+2n^2+xn+n +kx}{kn^2+knx+kn+n^3+xn^2+2n^2+nx+n} >1 $$

Thus the sequence is decreasing. ${\blacksquare}$

I find this approach extremely ugly (in particular the expansion of product). In simpler versions of the same problem i was almost always able to find some substitution which simplified the inequality greatly. So basically i have two questions:

1. Is my proof valid? (disregarding its ugliness)
2. What would be a smarter way to prove what's in the problem statement?

Please note the precalculus tag. Thank you!

Answer 1

You can do this:

$$1+\dfrac{x(n+k)}{n^2+nk+n}>1+\dfrac{x(n+k)}{n^2+nk+n+k} = 1+\dfrac{x(n+k)}{(n+1)(n+k)}>1+\dfrac{x}{n+1}.$$