Determine whether the functional series $\sum\frac1{(x+j)^2}$ and $(\sin\frac{x}{j})^j$ are pointwise and/or uniform convergent

Question

Determine whether the following functional series is pointwise and/or uniformly convergent

1) $\sum_{j=1}^{\infty} \frac{1}{(x+j)^2}$ with $(x>0)$

2)$\sum_{j=1}^{\infty} (\sin\frac{x}{j})^j$ with $(x>0)$

Im really unsure where to begin here?

For 1) $\sum_{j=1}^{\infty} \frac{1}{(x+j)^2} < \sum_{j=1}^{\infty}\frac{1}{j}$ which is divergent but I cant work out how this would help.

Struggling on what to compare them too! Help please

Answer 1

Hints: 1) $1/(x+j)^2 \le 1/j^2$ for all $x>0.$

2) Pointwise convergence: For any $x > 0$ there exists $j_0$ such that $0<\sin (x/j_0) <1/2.$ For all $j\ge j_0,$ the absolute value of the $j$th term is therefore $<(1/2)^j.$

Uniform convergence: The series converges uniformly on every $[0,b].$ This follows from the Weierstrass M-test and the idea above. However, the series does not converge uniformly on $[0,\infty).$ To see that, use the fact that if it did converge uniformly on $(0,\infty),$ then

$$\sup_{x\in (0,\infty)}|(\sin(x/j))^j|\to 0 \text { as } j\to \infty.$$

That fails here.