0
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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