So the question is from a textbook. I have spent some time thinking about this, but I do not think I have an adequate answer. It is the following:
Let $a_n$ and $b_n$ be two sequences such that the series $\sum^{\infty}_{n=1} a_n$ converges and $\lim_{n \to \infty}\frac{a_n}{b_n} = 1$. Both $a_n$ and $b_n$ are nonnegative.
a) Prove that the series $\sum^{\infty}_{n=1} b_n$ converges.
b) Prove that the Fourier series $\sum^{\infty}_{n=1} (a_n \cos n x + b_n \sin n x)$ converges uniformly and absolutely on $\mathbb{R}$.
The problem is that I am stumped on the first part. While it seems intuitively obvious that as $n$ approaches infinity, $b_n$ would have to converge in order for the fraction $\frac{a_n}{b_n}$ to converge, but how would one go about proving this rigorously?
As for the second part, is the answer a corollary to answering the first one? Any help would be much appreciated.
Edit: First the OP did not mention that the sequences $a_n$ and $b_n$ are nonnegative.
The statement is true. Since $a_n$ is nonnegative we have $|a_n|=a_n$ and therefore $\sum |a_n|$ is convergent.
Furthermore we have also $\frac{a_n}{b_n}\to 1$. So there exist an $N$ such that for all $n>N$ we have: \begin{align} |\frac{a_n}{b_n}-1| < \frac{1}{2} \end{align} So: \begin{align} \frac{1}{2}<\frac{a_n}{b_n} < \frac{3}{2}\\ \end{align} And that means: \begin{align} b_n < 2a_n \end{align} So $\sum b_n$ converges by the comparison test. Since it is nonnegative it also converges absolutely. For the Fourier series note that we have: \begin{align} \sum_{n=1}^\infty |(a_n\cos(nx) + b_n\sin(nx)) |\leq \sum_{n=1}^\infty (|a_n| + |b_n|) \end{align} Since the RHS is convergent and the inequality holds for all $x\in\mathbb{R}$ the Fourier series converges uniformly and absolutely by the Weierstrass-M test.