On a series of functions exercise: $\sum_{n=1}^{\infty} n \ln (1+ \frac{|\sin x|^n}{1+|x|^n})$. I want to study the uniform convergence of this series of functions:
$$\sum_{n=1}^{\infty} n \ln (1+ \frac{|\sin x|^n}{1+|x|^n})$$
After trying the root test on the sequence formed by moving the $n$ to the exponent of the argument of the logarithm I proceeded in this way:
I Mclaurin expanded $$\ln(1+\frac{|\sin x|^n}{1+|x|^n}) =  \frac{|\sin x|^n}{1+|x|^n} + \frac{1}{2}\Big(\frac{|\sin x|^n}{1+|x|^n} \Big)^2 + o(1/x) $$
Now one can see that if $x\le 1$ the series diverges (because we have this expansion multiplied by $n$), but what about the other values of $x$?
Edit: I added absolute values around x in the denominator and I can't Mclaurin expand when $\frac{|\sin x|^n}{1+|x|^n}$ does not go to zero so what I wrote above does not make sense.
 A: For all $x \ge 0$ we have $\ln(1+x) \le x$
Let $x > x_0> 1$ we have $\sum_{n=1}^{\infty} n \ln (1+ \frac{|sinx|^n}{1+x^n}) \le \sum_{n=1}^{\infty} n \ln (1+ \frac{1}{1+x_0^n})\le \sum_{n=1}^{\infty} n \frac{1}{1+x_0^n}$ and by the M-test of Weierstrass, the series converge uniformly.
A: I suppose that you are on $I=[0,+\infty($ (note that  for $x=-1$ and $n$ odd,  $1+x^n=0$). 
Put $\displaystyle v_n(x)=\frac{|\sin(x)|^n}{1+x^n}$.
a) First suppose that $\displaystyle 0\leq x\leq 1.5=u <\frac{\pi}{2}$ (hence $|\sin(u)|<1$). Then
$$0\leq v_n(x)\leq |\sin(x)|^n \leq \sin(u)^n=a_n$$
b) Suppose now that $x\geq u=1.5$. We have
$$0\leq v_n(x)\leq \frac{1}{1+u^n}\leq  u^{-n}=b_n$$
Now we have $\log(1+x)\leq x$ for all $x\geq 0$.
Hence for all $x\in I$ and $n$
$$u_n(x)=n\log(1+v_n(x))\leq nv_n(x)\leq n\sin(u)^n+nu^{-n}=na_n+nb_n=c_n$$
It remains to show that the series $c_n$ is convergent, it is easy.
A: Claim: The series converges absolutely, hence converges, for all $x\in \mathbb R.$ Note that because the terms are even in in $x,$ we only have to prove this for $x\ge 0.$
Two facts: i) $\ln (1+u) \le u$ for $u \ge 0;$ ii) $\sum_n nr^n < \infty$ for $0\le r <1.$ Using fact i), we have
$$\tag 1 n\ln (1 + \frac{|\sin x|^n}{1+|x|^n}) \le n\frac{|\sin x|^n}{1+|x|^n}$$
for all $n,x.$ Now for $x \notin \pi/2 +\pi\mathbb N,$ $|\sin x|<1,$ hence $(1)$ is less than or equal to $n|\sin x|^n.$ Fact ii) then implies our series converges absolutely. If $x \in \pi/2 + \pi\mathbb N,$ the right side of $(1)$ equals
$$n\,\frac{1}{1+|x|^n} \le n\frac{1}{(\pi/2)^n}.$$
Again using fact ii), we see our series converges absolutely. This proves the claim.
