| 
October 29th, 2007, 09:20 PM
|  | Global Moderator | | Join Date: Nov 2005 Location: New York City
Posts: 11,186
Country: Thanks: 482
Thanked 3,751 Times in 3,070 Posts
| | Problem 39 Let  . For each positive integer  let  .
Find the limit,
(Where  are possibly different integers defining a different sequence). | 
October 31st, 2007, 02:20 AM
|  | MHF Contributor | | Join Date: Oct 2007 Location: Taguig City, Philippines
Posts: 1,027
Country: Thanks: 72
Thanked 286 Times in 275 Posts
| | Quote:
Originally Posted by ThePerfectHacker Let  . For each positive integer  let  .
Find the limit,
(Where  are possibly different integers defining a different sequence). | is this lim as n approaches infinity? | 
October 31st, 2007, 07:45 AM
|  | Global Moderator | | Join Date: Nov 2005 Location: New York City
Posts: 11,186
Country: Thanks: 482
Thanked 3,751 Times in 3,070 Posts
| | Quote:
Originally Posted by kalagota is this lim as n approaches infinity? | When you are dealing with sequences that is the only type of limit you can have. (What else can you possibly approach  ) | 
October 31st, 2007, 10:26 AM
|  | Eater of Worlds | | Join Date: Jul 2006 Location: Chaneysville, PA
Posts: 2,853
Country: Thanks: 120
Thanked 1,098 Times in 986 Posts
| | Maybe I'm off base here, PH,:
So, we have: | 
October 31st, 2007, 11:09 AM
|  | Global Moderator | | Join Date: Nov 2005 Location: New York City
Posts: 11,186
Country: Thanks: 482
Thanked 3,751 Times in 3,070 Posts
| | Quote:
Originally Posted by galactus Maybe I'm off base here, PH,:  | The sums diverges. Because,  for sufficiently large  .
And  does not converge, it is the harmonic series. | 
November 1st, 2007, 07:41 AM
|  | MHF Contributor | | Join Date: Jun 2007 Location: Medgidia, Romania
Posts: 1,158
Country: Thanks: 22
Thanked 617 Times in 558 Posts
| | Let  .  is ascending and unbounded.
By Stolz-Cesaro, we have  .
Then | | The following users thank red_dog for this useful post: | |  | 
November 3rd, 2007, 10:07 PM
|  | Global Moderator | | Join Date: Nov 2005 Location: New York City
Posts: 11,186
Country: Thanks: 482
Thanked 3,751 Times in 3,070 Posts
| | Quote:
Originally Posted by red_dog Let  .  is ascending and unbounded.
By Stolz-Cesaro, we have  .
Then  | I can do it without Stolz-Cesaro. Want to try? | 
November 10th, 2007, 08:49 PM
|  | Global Moderator | | Join Date: Nov 2005 Location: New York City
Posts: 11,186
Country: Thanks: 482
Thanked 3,751 Times in 3,070 Posts
| | The key step is to note that  . That means,  . | | Thread Tools | | | | Display Modes | Linear Mode |
Posting Rules
| You may not post new threads You may not post replies You may not post attachments You may not edit your posts HTML code is Off | | | All times are GMT -7. The time now is 12:50 AM. | | |