How to show that $x_n=-\sqrt{n} + n\ln\Big(1+\frac{1}{\sqrt{n}}\Big)$ is decreasing? I am a non-mathematician who knows some elemententary calculus ans I want to prove that the sequence $(x_n)$ given by

$$
x_n=-\sqrt{n} + n\ln\Big(1+\frac{1}{\sqrt{n}}\Big)
$$

is decreasing. Is there an elegant way to show this?
This does not answer your question but might be helpful. Let $x&gt;0$ be a real number.  Then it follows from $$\log(x) = \int_1^x \frac{dt}{t}$$ and the estimates $$2-t &lt;\frac{1}{t}&lt;1-\frac{(t-1)x}{x+1}$$ for $t\in(1,1+1/x)$ that

$$
x-\frac{1}{2} &lt; x^2\log\left(1+\frac{1}{x}\right)&lt;x-\frac{1}{2}+\frac{1}{2(x+1)}.
$$

In particular $$\lim_{x\to\infty}x^2\log\left(1+\frac{1}{x}\right)-x=-\frac{1}{2}$$ which might or might not be what you are after.
as @user130512 said, you could prove that $x_{n+1}$ is greater than $x_n$. Sometimes, it can help to look at the ratio between the two, that is $x_{n+1}/x_n$. If it is greater than $0$, $x_{n+1}$ is greater than $x_n$.
I think you also should be able to look at the derivative of $x_n$ to deduce whether it is positive or negative. So,  positve $ \frac{d(x_n)}{dn}$ means it is increasing while negative would mean it is decreasing. Note that it can be increasing for one period but decreasing for another, so, as in your case, you would need to prove that $ \frac{d(x_n)}{dn}$ is smaller than $0$ for all values of $n$ to show that it is negative since the derivative will not be a constant. 


The derivative in your case is:

$$\dfrac{2(n+\sqrt{n})\log\left(\tfrac1{\sqrt{n}}+1\right)+1}{2(n+\sqrt{n})}$$

As you can see, we only need to find what values would make this function equal to zero. Because of this, we can work with the nominator only, $$(n+\sqrt{n})\times \log\left( \frac{1}{\sqrt{n}} +1\right)+1 =0$$ 

You can also see it graphically in the derivative: