Timeline for Show that $\sum c_n$ is not Cesaro summable
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 16, 2018 at 6:43 | answer | added | DanielWainfleet | timeline score: 0 | |
| Mar 16, 2018 at 6:31 | comment | added | Leyla Alkan | thanks for your time! | |
| Mar 16, 2018 at 6:28 | comment | added | dezdichado | never mind, I got it wrong. | |
| Mar 16, 2018 at 6:21 | vote | accept | Leyla Alkan | ||
| Mar 16, 2018 at 6:13 | answer | added | Kavi Rama Murthy | timeline score: 2 | |
| Mar 16, 2018 at 5:24 | comment | added | Leyla Alkan | @dezdichado This is how I do that: $\sigma_n=\frac {s_1+\cdots +s_n}{n}, \sigma_{n-1}=\frac {s_1+\cdots +s_{n-1}}{n-1} $ so, $\frac {s_1+\cdots +s_n}{n}-\frac{n-1}{n}\frac {s_1+\cdots +s_{n-1}}{n-1}=\frac {s_n}{n}$ How do you get $\frac {c_n}{n}$? | |
| Mar 16, 2018 at 5:17 | comment | added | dezdichado | you are mistaken. $\sigma_n - \dfrac{n-1}{n}\sigma_{n-1}$ is indeed equal to $\dfrac{c_n}{n}.$ Your challenge would be to prove that the left hand side tends to $0$, provided that $\sigma_n\to \sigma,$ some finite number. | |
| Mar 16, 2018 at 5:14 | comment | added | Leyla Alkan | Yeah, I definitely understand this, but I just didn't get what the book meant there @dezdichado | |
| Mar 16, 2018 at 5:12 | comment | added | dezdichado | Just look at $\dfrac{c_n}{n}$ and see if it tends to zero? | |
| Mar 16, 2018 at 5:08 | history | asked | Leyla Alkan | CC BY-SA 3.0 |