Timeline for Convergence of a series
Current License: CC BY-SA 3.0
17 events
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| Aug 8, 2016 at 7:15 | comment | added | Deepti | @Anton: Maybe, this is a bit shorter version of your proof. Set $S_0:=0,$ in addition to $S_N, N\geq 1$ defined previously. Correspondingly, $T_0= -ab_0.$ Then $\sum_{j=0}^Ma^jT_j=-a^{M+1}S_{M+1}.$ As $M\to \infty,$ then the left-hand side tends to $F(a)=0$ (this was asked by Alexandre Eremenko), whence $\{a^jS_j\}\to 0$ and we obtain $F_N=S_N.$ And then your reasoning on the Cauchy sequence. Thank you so much for your interest in the question and nice solution, Deepti. | |
| Aug 4, 2016 at 12:44 | history | edited | user1688 | CC BY-SA 3.0 |
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| Aug 4, 2016 at 12:38 | comment | added | user1688 | @Deepti: it does, however it can be repaired and I did so. | |
| Aug 4, 2016 at 12:37 | history | edited | user1688 | CC BY-SA 3.0 |
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| Aug 4, 2016 at 10:00 | comment | added | Deepti | @Anton: Does it matter that the limit $A$ depends on $N$? | |
| Aug 4, 2016 at 8:07 | history | edited | user1688 | CC BY-SA 3.0 |
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| Aug 4, 2016 at 7:56 | comment | added | user1688 | @Alexandre Eremenko: In the definition of $F(z)$. | |
| Aug 4, 2016 at 7:41 | comment | added | Alexandre Eremenko | @Anton: Where does your proof use that $F(a)=0$ ? It is clear that the statement does not hold without this assumption. | |
| Aug 4, 2016 at 7:01 | history | edited | user1688 | CC BY-SA 3.0 |
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| Aug 4, 2016 at 6:29 | comment | added | Deepti | I think there is a small misprint: it should be $S_N-a^{M+1}S_{N+M+1} $ in the formula for the sum $\sum_{j=0}^M a^jT_{N+j}.$ | |
| Aug 4, 2016 at 6:25 | vote | accept | Deepti | ||
| Aug 3, 2016 at 20:05 | history | edited | user1688 | CC BY-SA 3.0 |
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| Aug 3, 2016 at 17:05 | history | edited | user1688 | CC BY-SA 3.0 |
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| Aug 3, 2016 at 16:48 | history | undeleted | user1688 | ||
| Aug 3, 2016 at 16:48 | history | edited | user1688 | CC BY-SA 3.0 |
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| Aug 3, 2016 at 16:15 | history | deleted | user1688 | via Vote | |
| Aug 3, 2016 at 16:10 | history | answered | user1688 | CC BY-SA 3.0 |