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Consider this modified version of the harmonic series:

$\sum_{n=1}^\infty \frac{\varepsilon_n}{n}$

With

${\varepsilon_k}=\begin{cases} 0, & \text{if $k$ ends in 9 in its decimal representation} \\ 1, & \text{otherwise} \end{cases}$

This series diverges. I tried to prove this like follows:

I rewrited the original series in this way:

$\sum_{n=1}^\infty (\frac{1}{10n} + \frac{1}{10n-2}+ \frac{1}{10n-3}+ \frac{1}{10n-4}+ \frac{1}{10n-5}+ \frac{1}{10n-6}+ \frac{1}{10n-7}+ \frac{1}{10n-8}+ \frac{1}{10n-9})$

Now it is easy to show that, for example, $\sum_{n=1}^\infty \frac{1}{10n}$ diverges, so the whole series diverges. Is this enough to prove the divergence of the original series? The new series is not strictly the same as the original, but it has the same terms. Maybe I am missing a step here to provide full rigourosity.

I would like to know your opinion about correctness/rigourosity of this proof, and any other possible idea to prove the series diverges is also welcome!

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  • $\begingroup$ You are basically correct! The key detail here is that all these series have nonnegative terms. That means there's only two things that can happen: convergence, or divergence to $+\infty$. And the Comparison Test applies, so that the divergence of the smaller series $\sum_{n=1}^\infty \frac1{10n}$ does imply the divergence of the larger series $\sum_{n=1}^\infty \frac{\epsilon_n}n$. $\endgroup$ – Greg Martin Sep 1 '16 at 19:27
  • $\begingroup$ It looks better adding that argument. Thanks! $\endgroup$ – bowman_d Sep 1 '16 at 20:51

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