For which $p \in [0, +\infty)$ $\left(\log\left(1 + \frac{e^{px}}{e^x + n}\right)\right)_{n}$ converges uniformly to $0$ on $\mathbb{R}_+$? The question is to find  for which $p \geq 0$ the sequence of functions $$f_n: x\in\mathbb{R}_+ \mapsto \log\left(1 + \frac{e^{px}}{e^x + n}\right)$$ converges uniformly to $0$ on $\mathbb{R}_+$ ?
I want to say that for $p \leq 1$ (because we want at least pointwice convergence to $0$). But what's the correct answer for this and how can we prove it?
I know Abel's and Dirichlet's test, but seem's like they're not working here.
Also I found that $\log(1 + x) \leq x$ and seems like we only have to find $p$ such that $\dfrac{e^{px}}{e^x + n}$ converges uniformly to $0$.
 A: For $p>1$, $\dfrac{e^{px}}{e^{x}+n}\rightarrow +\infty$ therefore you can easily see that $f_n$ is not bounded and therefore $(f_n)_{n\in\mathbb{N}}$ does not converge uniformly to $0$ on $\mathbb{R}_+$.

For $p = 1$, let's determine $||f_n||_{\infty}$ for all $n \in \mathbb{N}$. Let $n \in \mathbb{N}$.
We have for all $x\geq 0$,$f_n'(x)=\dfrac{n e^{x}}{(e^x+n)(2e^x+n)}\geq0$ therefore $f_n$ is increasing on $\mathbb{R}_+$
$\displaystyle \lim_{n\rightarrow\infty} f_n(x) = \log(2) $ therefore $\lVert f_n \rVert_{\infty}= \log(2) $ thus $\displaystyle \lim_{n\rightarrow +\infty} ||f_n||_{\infty} = \log(2)\neq0$ and therefore $(f_n)_{n\in\mathbb{N}}$ does not converge uniformly on $\mathbb{R}_+$ to $0$.

For $0 \leq p<1$ we have,
$$f_n'(x)=\dfrac{e^{px}((p-1)e^x+np)}{(e^x+n)(e^{px}+e^x+n)}$$

So if $0<p<1$,
$$f_n'(x) = 0\iff(1-p)e^x = np \iff x= \log\left(\dfrac{np}{1-p}\right)$$
And you can easily check it is a maximum (the derivative is firstly positive then negative). Therefore,
$$||f_n||_\infty=f_n\left(\log\left(\dfrac{np}{1-p}\right) \right) = \log\left(1+\dfrac{\left(\frac{np}{1-p}\right)^p}{\frac{np}{1-p}+n} \right)$$
But,
$$\dfrac{\left(\frac{np}{1-p}\right)^p}{\frac{np}{1-p}+n} \underset{n \rightarrow \infty}{\sim} n^{p-1} \left( \frac{\left(\frac{p}{1-p}\right)^p}{1+\frac{p}{1-p}}\right) \underset{n \rightarrow \infty}{\sim} \alpha_p \dfrac{1}{n^{1-p}} \underset{n \rightarrow \infty}{\longrightarrow} 0 $$
Therefore,
$$ ||f_n|| \underset{n \rightarrow \infty}{\rightarrow} \log(1+0) =0$$
So for $0<p<1$,  $(f_n)_{n\in\mathbb{N}}$ converges uniformly on $\mathbb{R}_+$ to $0$.

Finally for $p=0$, we have, by using $|\log(1+x)|\leq |x|$
$$\forall n\in \mathbb{N},\forall x\geq0,|f_n(x)|\leq \dfrac{1}{e^x+n} \leq \dfrac{1}{1+n}$$
Therefore we deduce that,
$$\forall n\in \mathbb{N}, ||f_n||_{\infty} \leq \dfrac{1}{1+n}$$
Hence,
$$ ||f_n|| \underset{n \rightarrow \infty}{\rightarrow} 0$$
So for $p=0$,  $(f_n)_{n\in\mathbb{N}}$ converges uniformly on $\mathbb{R}_+$ to $0$.

Conclusion:
$(f_n)_{n\in\mathbb{N}}$ converges uniformly on $\mathbb{R}_+$ to $0$ if and only if $0\leq p <1$.
