If a new series is formed by taking every $n\mathrm{th}$ term (eg for $n=3$, the 3rd, 6th, 9th etc terms) of the harmonic series$$\sum_{k=1}^{\infty}\frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots$$ is this new series divergent?

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    $\begingroup$ Yes it is. For example, for $n=3$, you have $$\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12} +\cdots = \frac{1}{3} \left( \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \cdots \right) = \infty $$ $\endgroup$ – Crostul Mar 27 at 14:59
  • $\begingroup$ Whoops, I should have seen that coming. If you make that comment an answer, I'll happily accept it :-) $\endgroup$ – Peter4075 Mar 27 at 15:05

Yes, it is. For example, for $n=3$, you have $$\frac{1}{3}+ \frac{1}{6}+ \frac{1}{9}+ \frac{1}{12}+ \cdots = \frac{1}{3} \left( \frac{1}{1}+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \cdots\right) = \infty$$


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