Timeline for Convergence of $\sum_{n=1}^\infty x_n^k$
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 25, 2020 at 4:49 | comment | added | metamorphy | A solution is here. | |
| Oct 10, 2019 at 9:57 | comment | added | Gerry Myerson | So, on some website, you were personally asked to prove this? | |
| Oct 2, 2019 at 13:34 | comment | added | Ma Joad | @GerryMyerson I cannot remember exactly. I saw it about a year ago on some website, which I cannot remember. It is NOT my homework. | |
| Oct 2, 2019 at 13:21 | comment | added | Gerry Myerson | "I am asked to prove..." Who asked you to prove this, and under what circumstances, please? | |
| Oct 2, 2019 at 10:42 | comment | added | Anthony Quas | This guarantees convergence (to 0) for every odd power below $2k+1$; divergence for $2k+1$ and convergence for everything above. Now find a way to mix these sequences. | |
| Oct 2, 2019 at 10:41 | comment | added | Anthony Quas | From the way you ask, this sounds like a homework question (which as you guessed is discouraged here). I will suggest a hint (which I haven't completely worked out, but I think will work). For each odd number, $2k+1$, here's a suggested way to build a sequence which diverges for that power alone. Find a finite collection of numbers $T_k=\{a_1,\ldots,a_l\}$ such that $\sum_{a\in T_k}a^{2j+1}=0$ for $j=1,\ldots,k-1$ and $\sum_{a\in T_k}a^{2j+1}\ne 0$ for $j\ge k$ (e.g. Prouhet). Then build a sequence by concatenating the block $a_12^{-j},\ldots,a_l2^{-j}$ repeated $2^{(2k+1)j}$ times over. | |
| Oct 2, 2019 at 9:49 | history | asked | Ma Joad | CC BY-SA 4.0 |