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.

  • $\begingroup$ 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

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 $\alpha <1$.


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 $\sum_{k=0}^{m} \frac{s_k}{k} \geq \sum_{n=0}^{m} \frac{1}{n}$


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