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.

  • $\begingroup$ Is $a_n$ absolutely convergent? $\endgroup$ – copper.hat Nov 17 '17 at 6:55
  • $\begingroup$ It is only given that it converges, absolute convergence cannot be assumed. $\endgroup$ – user485755 Nov 17 '17 at 6:57
  • $\begingroup$ Okay does it say that the sequences are nonnegative? Because as it is written there is not true see my answer below. $\endgroup$ – Shashi Nov 17 '17 at 10:12
  • $\begingroup$ If $a_n \ge 0$ then the sequence is absolutely convergent!!! $\endgroup$ – copper.hat Nov 17 '17 at 16:54
  • $\begingroup$ Note, with $a_n = (-1)^n {1 \over n}$ and $b_n = a_n + {1 \over n \log n}$ we have $\sum_n a_n$ conditionally convergent, ${a _n \over b_b} \to 1$ and $\sum_n b_n $ divergent. $\endgroup$ – copper.hat Nov 17 '17 at 18:23

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.

  • $\begingroup$ I forgot to put in the statement saying that they are both non-negative! How does that change your answer? $\endgroup$ – user485755 Nov 17 '17 at 16:07
  • $\begingroup$ @Mathmaniac the answer is totally different. I deleted the last one and added this one. I hope it helps you. $\endgroup$ – Shashi Nov 17 '17 at 16:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy