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Prove/Disprove that if $\sum a_n$ and $\sum b_n$ are some series and $\underset{n\to \infty }{\mathop{\lim }}\,{{a}_{n}/b_n } = 1$ then the series converge together or not converge together.

This doesn't seem to be correct to me so maybe there is some counter example. i know this is true if both series are strictly nonnegative (from the first series comparison test) - how does it change if both are negative (or alternate?)

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marked as duplicate by Chinnapparaj R, Cm7F7Bb, Community Dec 5 '18 at 13:10

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  • $\begingroup$ This is true for series with a posiitve (or negative) general term. It is the criterion by equivalence of functions. $\endgroup$ – Bernard Dec 5 '18 at 13:00
  • $\begingroup$ Thank you very much everyone for your time, turns out it was a duplicate.. $\endgroup$ – Boaz Yakubov Dec 5 '18 at 13:11
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If $\underset{n\to \infty }{\mathop{\lim }}\,{{a}_{n}/b_n } = 1$ , then $$(\exists N\in \Bbb N, \forall n > N):\; \frac{1}{2}a_n \le b_n\le \frac{3}{2}a_n$$

Thus if $\sum a_n$ converges then so does $\sum b_n$ and vice versa

Update As stated in the comment this is infact only true if $a_n$ and $b_n$ keep a constant sign (positive or negative) after a certain n. Otherwise it's false: cf. the counter example given below

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    $\begingroup$ @A.Pongrácz For series of positive numbers this is a complete proof. In the example you give, both series diverge. $\endgroup$ – MaoWao Dec 5 '18 at 13:09
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A counterexample is: $$a_n = \frac{(-1)^n}{\sqrt{n}} \text{ and } b_n = \frac{(-1)^n}{\sqrt{n} + (-1)^n}.$$

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