Timeline for Growth of a particular sequence
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2012 at 13:42 | history | edited | Igor Rivin | CC BY-SA 3.0 |
fixed constant, as per @Barry, introduced clarification
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| Jun 15, 2012 at 13:41 | comment | added | Igor Rivin | @Barry: yes, you are right, I did not notice the start index. @Gerry: I guess what I had meant to say is that most "closed form" numbers known to be transcendental have names, and it is unlikely that this is one of them... | |
| Jun 15, 2012 at 13:13 | comment | added | Barry Cipra | Igor, I think your compact expression should have a 4 out in front instead of a 2 -- the OP's term for $k=3$, for example, is $3^3/2^{2+4} = 3^3/2^{2^3-2}$. The problem is that the product index should really start with $i=0$ in order for the $k=1$ term to make proper sense -- if it did, then your expression would match. | |
| Jun 15, 2012 at 2:00 | vote | accept | Turbo | ||
| Jun 15, 2012 at 2:13 | |||||
| Jun 15, 2012 at 1:53 | vote | accept | Turbo | ||
| Jun 15, 2012 at 1:53 | |||||
| Jun 15, 2012 at 1:50 | comment | added | Turbo | Basically Gerry's and Igor's answer says it converges to a constant. ok. | |
| Jun 15, 2012 at 1:16 | comment | added | Turbo | $S_{m} −S_{m−1} = \frac{3^{m}}{2^{2^{m}−1}}$ | |
| Jun 15, 2012 at 0:23 | comment | added | François G. Dorais | The sequence $S_m$ converges to some number between $2.83$ and $2.84$, so it doesn't grow very much... | |
| Jun 14, 2012 at 23:51 | comment | added | Turbo | Hi Igor, I am only interested in something that grows as fast as $S_{m}$ and not S_{m}$ accurately. | |
| Jun 14, 2012 at 22:56 | comment | added | Gerry Myerson | $\sum_{n=0}^{\infty}(1/n!)$ is a lacunary sum with a transcendental value, yet it has what would generally be accepted as a closed form; I'm not objecting to your conclusion, just to the word, "so". | |
| Jun 14, 2012 at 22:15 | history | answered | Igor Rivin | CC BY-SA 3.0 |