Show if the series $\sum _{n=1}^{\infty} \frac {3}{9n+1}$ converges or diverges.

The solution says that by comparison test with harmonic series this series diverges. But, $\frac {1}{n}$ is larger than $\frac {1}{9n+1}$. That is, the divergence of harmonic series does not tell us whether $\sum _{n=1}^{\infty} \frac {3}{9n+1}$ converges or not.

Am I correct? Could you give me the alternative to solve this problem?

  • 1
    $\begingroup$ You should indeed do a test against a harmonic series, but not necessarily one with terms $\frac{1}{n}$, you could also do it against ones with terms $\frac{1}{3n}$ or so, remembering that $\sum\frac{1}{3n}=\frac{1}{3}\sum\frac{1}{n}$. So... lets try to rearrange things in a useful manner... if we want to compare to something smaller in a way to get rid of the $+1$ on the bottom lets make the $+1$ something larger... like... how about $n$... $\endgroup$
    – JMoravitz
    Commented Jun 18, 2018 at 2:18
  • 1
    $\begingroup$ Compare with $\frac{3}{9n+9n}$ which is indeed smaller $\endgroup$
    – user223391
    Commented Jun 18, 2018 at 2:19

1 Answer 1


\begin{align} \sum_{n=1}^{\infty}\frac{3}{9n+1} &= \frac{1}{3}\sum_{n=1}^{\infty}\frac{1}{n+1/9}\\ &=\frac{1}{3}\sum_{n=2}^{\infty}\frac{1}{n-8/9} \gt\frac{1}{3}\sum_{n=2}^{\infty}\frac{1}{n}\\ \text{And the last term diverges by harmonic series} \end{align}


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