0
$\begingroup$

I recently thought of this problem (though I by no means think others haven't...) and a couple of solutions; I figured I would share. I think this is an interesting problem for those learning analysis.

Prove or provide a counterexample: Let $\{x_n\}$ be a sequence in $\mathbb{R}$ that converges to $0$. Then $\sum_{n=0}^{\infty} \frac{x_n}{n}$ is a convergent series.

I think this is interesting because it tackles a boundary case of "how quickly must a series go to 0", as for any $\epsilon >0$, $\sum_{n=0}^{\infty} \frac{x_n}{n^{1+\epsilon}}$ converges. This is a nice situation of "can you find a sequence that converges slowly enough."

I'd also be interested to see other counter-examples.

$\endgroup$
  • $\begingroup$ Convergence of $\sum x_n/n$ equivavent to convergence of $\sum x_n$, and condition $x_n\to0$ is neccesary, but not sufficient. $\endgroup$ – Michael Galuza Jun 8 '15 at 5:32
  • 2
    $\begingroup$ No, it is not equivalent. $\endgroup$ – Robert Israel Jun 8 '15 at 6:14
  • $\begingroup$ In this line of thought you'd like to read these links which discuss the nonexistence of a boundary between convergent and divergent series: (1)math.stackexchange.com/questions/588488/… (2) mathoverflow.net/questions/49415/… $\endgroup$ – hjhjhj57 Jun 8 '15 at 6:28
  • $\begingroup$ @MichaelGaluza That is false; take $x_n$ = $\frac{1}{n}$ $\endgroup$ – Layne Jun 9 '15 at 18:22
  • $\begingroup$ @Layne, yes, i realize it. I used root test for both series and $\lim \sqrt[n]{x_n} = lim \sqrt[n]{x_n/n}$. My bad. $\endgroup$ – Michael Galuza Jun 9 '15 at 19:07
1
$\begingroup$

What follows is relevant:

  1. Let $(u_n)$ be a decreasing sequence that goes to $0$. If $\sum u_n$ converges, then $u_n=o(\frac{1}{n})$

  2. There is no borderline divergent series with positive terms: let $u_n$ be a positive sequence such that $\sum u_n$ diverges. Let $S_n$ denote its partial sums. Then$\sum \frac{u_n}{S_n}$ diverges.

  3. There is no borderline convergent series with positive terms: let $u_n$ be a positive sequence such that $\sum u_n$ converges. Let $R_n=(\sum_{k=0}^\infty u_k)-S_n$. Then$\sum \frac{u_n}{R_n^\alpha}$ converges for every $\alpha <1$.

$\endgroup$
0
$\begingroup$

I have 2 solutions:

  1. Simple counterexample: $\{x_n\} = \frac{1}{log(n)}$. Then by the integral test, $\int\frac{1}{nlog(n)} = log(log(n))$ which increases monotonically to infinity.

  2. If you are going through Rudin in order and don't yet "know" what an integral is, the harmonic series diverges because for $\{\frac{1}{n}\}$, you can group finite blocks $s_k$ of terms into partial sums of 1 no matter how large n is, so one can construct a sequence $\{x_n\}$ that is equal to $\frac{1}{n}$ on the $n$'th block, and then for infinitely many $m$, we have $\sum_{k=0}^{m} \frac{s_k}{k} \geq \sum_{n=0}^{m} \frac{1}{n}$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.