Uniform convergence Check

Question

Question is to check :

I guess i am through with $(c)$ i.e., i see it as an infinite geometric series and say that :

$$x^2+\frac{x^2}{1+x^2}+\frac{x^2}{(1+x^2)^2}+\dots = \frac{x^2}{1-\frac{1}{1+x^2}}=1+x^2$$

So, this infinite series actually converges and is actually finite in given interval..

So, can i just say this converges "Uniformly"? I guess i can say.. but i am not sure..

I tried to somehow make second series interms of just exponential popwers but then i ruined it. (as you can see in previous form of this post)

I have no idea about first question...

I would be thankful if some one can help me with this Question.

Thank You

EDIT : Thanks to Mr.Ayman for suggesting Weierstrass Test , I tried some thing in that way :

$$|a_n \cos nt +b_n \sin nt|\leq |a_n||\cos nt| +|b_n| |\sin nt|\leq |a_n|+|b_n|$$

I only know $\sum _{n} |a_n| < \infty$ and $\sum _{n} |b_n| < \infty$ and i guess this would imply $$\sum _{n} |a_n|+|b_n|<\infty$$ which is sufficient to deal with case $(a)..$ but i am not so sure how to write this explicitly..

please help me in this case..

Thank You

Answer 1

Weirstrass M-Test is saying If $|f_n|\le M_n\forall n$ and $\sum M_n$ is convergent Then $\sum f_n$ is uniformly convergent. Note That Converse is not true.

We will apply this result to solve $a.$

where $f_n(t)=a_n\sin nt+b_n\cos nt$, clearly $|f_n|\le a_n+b_n=M_n$ as $\sum (a_n+b_n)$ is given convergent so $\sum f_n=a_n\sin nt+b_n\cos nt$ is uniformly convergent

for $b$ you have done! and it is not UC

$c$ use the fact that if the sequence of continuous function converges uniformly to a function then the limit is also continuos

so only $a$ is true!