Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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. – Pete L. Clark Aug 7 '10 at 0:30
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. – Larry Wang Aug 7 '10 at 1:08
@Pete, I have no idea... was a long time ago- I just remembered the question out of the blue. – Tom Boardman Aug 7 '10 at 9:25
up vote 12 down vote accepted

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 $u=\log x$.

share|cite|improve this answer
I suggest using the condensation test (, which is easier than integrating 1/log(x), IMO – Tomer Vromen Aug 6 '10 at 23:33
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 $u=\log x$. – Samuel Aug 6 '10 at 23:49
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. – Tom Boardman Aug 7 '10 at 9:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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