Timeline for Deriving an asymptotic statement from a recursion
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 4, 2021 at 21:00 | comment | added | Iosif Pinelis | @მამუკაჯიბლაძე : Thank you for your comment. | |
| Aug 4, 2021 at 18:25 | comment | added | მამუკა ჯიბლაძე | MO at its best! | |
| Jul 6, 2021 at 14:19 | comment | added | Iosif Pinelis | @Testname420 : I have added a detail on that.: one also may note that $c^{-h}_k\to\frac{1-h}b<\infty$ as $k\to\infty$. | |
| Jul 6, 2021 at 14:17 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 6, 2021 at 10:33 | comment | added | Testname420 | I actually still have one question. I don't quite see how to use (2) and that $\prod_{j=2}^\infty (1+\frac{h}{j-1})=\infty$ to follow that (3) cannot hold forever. I see that using a telescopic product we can use $\prod_{j=2}^\infty (1+\frac{h}{j-1})=\infty$ to get that $c_n\to\infty$ if $m=m_n\geq n$ would not exist. But I don't know what to do next, or why this is a problem | |
| Jul 5, 2021 at 12:53 | comment | added | Testname420 | Thanks a lot for your time and your proof! | |
| Jul 5, 2021 at 12:44 | vote | accept | Testname420 | ||
| Jul 4, 2021 at 19:48 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 4, 2021 at 18:04 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 4, 2021 at 17:42 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 4, 2021 at 17:32 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 4, 2021 at 16:17 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 4, 2021 at 16:10 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |