Question: > 
> Does the following series converge uniformly in $[1,\infty)$? $$\sum\_{n=1}^\infty \frac{x^n}{(1+x)(1+x^2)\cdots (1+x^n)}$$
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I thought I might be able to prove it using [Weierstrass criterion (for uniform convergence)](https://encyclopediaofmath.org/index.php/Weierstrass_criterion_(for_uniform_convergence)), but all I go to was a lower bound of $1^n$ which does not converge.

Answer: 
We can suppose $n\geq 2$. Since $x^n/(1+x^n)\leq 1$ and $1+x^m \geq 2$ $\forall m \geq 0$, we have

$$
\frac{x^n}{(1+x)(1+x^2)\cdots (1+x^n)} \leq \frac{1}{(1+x)\cdots(1+x^{n-1})} \leq \frac{1}{2^{n-1}}
$$

and the result is clear by using Weierstrass M-Test.
Since Rafael Gonzalez Lopez already provided a proof for uniform convergence, I would like to note that the series converges uniformly to $1$ on $[1,\infty)$. To prove this, observe that
$$\frac{x^n}{\prod\_{j=1}^n\,\left(1+x^j\right)}=\frac{1}{\prod\_{j=1}^{n-1}\,\left(1+x^j\right)}-\frac{1}{\prod\_{j=1}^n\,\left(1+x^j\right)}\text{ for each }x>0\,.$$
That is, for each $m=1,2,3,\ldots$, we have
$$\sum\_{n=1}^m\,\frac{x^n}{\prod\_{j=1}^n\,\left(1+x^j\right)}=1-\frac{1}{\prod\_{j=1}^m\,\left(1+x^j\right)}\,.$$
The claim follows immediately.