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I am trying to make sense of the comparison test and I follow the proof for the comparison test for the convergence then the author appears to use the contra-positive to show if you flip the statements around and show that one diverges then the other must.

But that only makes sense to me if I assume that if a series does not converge then it must diverge . Is this true? Thanx.

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closed as off-topic by Andrés E. Caicedo, Rohan, José Carlos Santos, Namaste, user223391 Dec 28 '17 at 2:40

This question appears to be off-topic. The users who voted to close gave these specific reasons:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Rohan, Namaste, Community
  • "This question is not about mathematics, within the scope defined in the help center." – Andrés E. Caicedo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ It is true by definition. $\endgroup$ – Andrés E. Caicedo Dec 23 '17 at 16:03
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    $\begingroup$ People use "diverge" in two related but fundamentally different ways. I believe that standard usage is to say that any series which does not converge must diverge. However, people often distinguish between a series for which the partial sums go to $\infty$ and a series for which the partial sums oscillate, as in $\sum (-1)^n$. You should clarify what you mean by "diverges". $\endgroup$ – lulu Dec 23 '17 at 16:03
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    $\begingroup$ You find definitions in textbooks. Thank you. $\endgroup$ – Professor Vector Dec 23 '17 at 16:04
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    $\begingroup$ I've always taken "diverges" to mean "does not converge". There's also a concept of "diverging to" something, such as $+\infty$ or $-\infty$ or (in the case of infinite products) $0. \qquad$ $\endgroup$ – Michael Hardy Dec 23 '17 at 16:59
  • $\begingroup$ your comments have answered the question.....it is by definition and convention ....thank you $\endgroup$ – Sedumjoy Dec 24 '17 at 16:35