This question just reminded me of a conundrum I posed myself in my first year of university. I never did get a satisfactory answer...

Let $a_n$ be a null sequence. Does it follow that $\sum \frac{a_n}{n}$ converges?

Any ideas?

  • 1
    $\begingroup$ Just curious: if you couldn't figure this out for yourself, why didn't you ask your instructor? It's what we're here for, you know. $\endgroup$ Commented Aug 7, 2010 at 0:30
  • 1
    $\begingroup$ I've added a link for definition of a null sequence, since Samuel's answer indicates not everyone is completely familiar with the term. I'd also like the question title to be more descriptive, but @Tom, you should choose one you feel most closely aligns with the intent of your question. $\endgroup$
    – Larry Wang
    Commented Aug 7, 2010 at 1:08
  • $\begingroup$ @Pete, I have no idea... was a long time ago- I just remembered the question out of the blue. $\endgroup$ Commented Aug 7, 2010 at 9:25

1 Answer 1


If by null sequence you mean a sequence that converges to 0, then no. Try $a_n=1/\log n.$ By integral comparison, the series diverges:

$$\sum_2^\infty\dfrac1{n\log n}\geq\int_2^\infty\dfrac{dx}{x\log x}=\int_{\log 2}^\infty\dfrac{du}u=\infty,$$ where I've used the change of variables $u=\log x$.

  • 3
    $\begingroup$ I suggest using the condensation test (en.wikipedia.org/wiki/Cauchy%27s_condensation_test), which is easier than integrating 1/log(x), IMO $\endgroup$ Commented Aug 6, 2010 at 23:33
  • 2
    $\begingroup$ The integral we get is $\int \dfrac{dx}{x\log x}$, which is transformed to $\int\dfrac{du}u=\infty$ by the change of variables $u=\log x$. $\endgroup$
    – Samuel
    Commented Aug 6, 2010 at 23:49
  • $\begingroup$ thanks. I'm sure my younger self could have sworn \sum 1/nlog(n) converged. In fact I'm sure this was the motivating example for me. As it happened, figured another counterexample as soon as I went to bed. $\endgroup$ Commented Aug 7, 2010 at 9:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .