Template talk:Series (mathematics)

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Alternate organization

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Hyacinth (talk) 02:00, 11 September 2012 (UTC)[reply]__DTELLIPSISBUTTON__{"threadItem":{"timestamp":"2012-09-11T02:00:00.000Z","author":"Hyacinth","type":"comment","level":1,"id":"c-Hyacinth-2012-09-11T02:00:00.000Z-Alternate_organization","replies":[]}}-->

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Harmonic series?

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Why isn't it in the box? Doops | talk 04:26, 15 January 2014 (UTC)[reply]__DTELLIPSISBUTTON__{"threadItem":{"timestamp":"2014-01-15T04:26:00.000Z","author":"Doops","type":"comment","level":1,"id":"c-Doops-2014-01-15T04:26:00.000Z-Harmonic_series?","replies":["c-Hyacinth-2014-07-24T21:12:00.000Z-Doops-2014-01-15T04:26:00.000Z"]}}-->

Added. Hyacinth (talk) 21:12, 24 July 2014 (UTC)[reply]__DTELLIPSISBUTTON__{"threadItem":{"timestamp":"2014-07-24T21:12:00.000Z","author":"Hyacinth","type":"comment","level":2,"id":"c-Hyacinth-2014-07-24T21:12:00.000Z-Doops-2014-01-15T04:26:00.000Z","replies":[]}}-->

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Riemann zeta function

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This series is listed as convergent, however, its convergence depends on what value is used for $s$ . —Eli355 (talk | contribs) 00:44, 12 July 2018 (UTC)[reply]__DTELLIPSISBUTTON__{"threadItem":{"timestamp":"2018-07-12T00:44:00.000Z","author":"Eli355","type":"comment","level":1,"id":"c-Eli355-2018-07-12T00:44:00.000Z-Riemann_zeta_function","replies":["c-D.Lazard-2018-07-12T06:58:00.000Z-Eli355-2018-07-12T00:44:00.000Z"]}}-->

It is common that a series that depends on a variable is said convergent if its domain of convergence is not empty. The problem is thus not in the template, but in Domain of convergence where only the case of power series is considered, and this (ab)use of terminology is not mentioned. D.Lazard (talk) 06:58, 12 July 2018 (UTC)[reply]__DTELLIPSISBUTTON__{"threadItem":{"timestamp":"2018-07-12T06:58:00.000Z","author":"D.Lazard","type":"comment","level":2,"id":"c-D.Lazard-2018-07-12T06:58:00.000Z-Eli355-2018-07-12T00:44:00.000Z","replies":[]}}-->