<s>[INST] How to manipulate the Algebra in the parenthesis of this question? Consider the series $1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \frac{1}{9} + ....$
Since the pattern of signs repeats every 3 terms, it is natural to consider only the partial sums that include entire blocks of 3 terms, i.e.,
$s_{3n} = \sum_{k=1}^{n} (\frac{1}{3k - 2} + \frac{1}{3k-1} - \frac{1}{3k})$
a) Use a comparison test against the Harmonic Series to show that $(s_{3n})$ diverges. To this end, rewrite the expression in parenthesis above appropriately.
Now from my understanding, we have to show that $(\frac{1}{3k - 2} + \frac{1}{3k-1} - \frac{1}{3k}) = \frac{C}{k} + a_{k}$ where $a_{k}$ is a positive sequence and C is a positive constant or sequence of positive constants, therefore, our original sequence bounds the sequence of partials sums of the harmonic series and we conclude by the Comparison test that the original diverges. If I am right, how would one go about doing this?
 [/INST] $$s_{3n} = \sum_{k=1}^{n} \left(\frac{1}{3k - 2} + \frac{1}{3k-1} - \frac{1}{3k}\right)\geqslant \sum_{k=1}^{n}\frac{1}{3k - 2}$$
as last diverged, then it have limit $+\infty$, so have $s_{3n}$.
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