# More convergent series

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?

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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

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 $u=\log x$.

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I suggest using the condensation test (en.wikipedia.org/wiki/Cauchy%27s_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
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