I'm having trouble showing whether or not the following series converges: $$\frac{1}{1}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}...$$ I've tried using several methods, such as creating two leibniz series from the terms, or using comparison tests, but failed to prove anything. Does this series converge and why? Thank you.
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5$\begingroup$ $$(\frac{1}{2}-\frac{1}{3})+(\frac{1}{5}-\frac{1}{6})+(\frac{1}{8}-\frac{1}{9})+...$$ is convergent and the remaining part is divergent . Hence the series is divergent. $\endgroup$– Kavi Rama MurthyCommented Jan 15, 2021 at 11:55
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2 Answers
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Hint
Ignore all but the 1st, 4th, 7th, 10th, ... terms, with the idea that the remainder of the terms (i.e. the ones being ignored) have a positive sum. Then, compare the terms not being ignored to (1/3) of the harmonic series, which is known to be divergent.
answered Jan 15, 2021 at 11:53
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$\begingroup$ Thank you! I completely missed the fact that this sum is (of the 2 sums) is divergent $\endgroup$– Omer PazCommented Jan 15, 2021 at 12:09
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The isolating of the first, fourth etc terms is the first thing I thought of, as in another proposed answer.
Another way is to group the terms other than the first in threes and study $$\frac 1{3n-1}-\frac 1{3n}+\frac 1{3n+1}$$ which (from the curvature of the reciprocal function I expect to be greater than $\dfrac 1{3n}$ which will do for comparison - indeed even if close to $3n$ a comparison is likely to be possible).
answered Jan 15, 2021 at 12:00