This is the current revision of this page, as edited by Skeptos(talk | contribs) at 14:04, 2 June 2025(In the last statement of the proof, there was a typo and a_n would be swapped with b_n to make the argument work. However, the proof could be improved.). The present address (URL) is a permanent link to this version.Revision as of 14:04, 2 June 2025 by Skeptos(talk | contribs)(In the last statement of the proof, there was a typo and a_n would be swapped with b_n to make the argument work. However, the proof could be improved.)
Let and for all natural numbers . Now
does not exist, so we cannot apply the standard comparison test. However,
and since converges, the one-sided comparison test implies that converges.
Let for all . If diverges and converges, then necessarily
, that is,
. The essential content here is that in some sense the numbers are larger than the numbers .
Let be analytic in the unit disc and have image of finite area. By Parseval's formula the area of the image of is proportional to . Moreover,
diverges. Therefore, by the converse of the comparison test, we have
, that is,
.
Rinaldo B. Schinazi: From Calculus to Analysis. Springer, 2011, ISBN9780817682897, pp. 50
Michele Longo and Vincenzo Valori: The Comparison Test: Not Just for Nonnegative Series. Mathematics Magazine, Vol. 79, No. 3 (Jun., 2006), pp. 205–210 (JSTOR)
J. Marshall Ash: The Limit Comparison Test Needs Positivity. Mathematics Magazine, Vol. 85, No. 5 (December 2012), pp. 374–375 (JSTOR)