Re: Solve This Sequences Problem
- From: fiazidris@xxxxxxxxx
- Date: 30 Aug 2006 20:50:06 -0700
Thank you for all.
Sorry, I wanted n > 1. I thought it was obvious, my mistake.
So, how does one find the solution n > 58 using minimal mathematics.
mensanator@xxxxxxxxxxx wrote:
fiazidris@xxxxxxxxx wrote:
1) Can someone write the steps for proving the following:
Prove 2^n > n^10 for large n
n = 1 isn't large.
2) Also, 2^n = n^10 , find n
I want to see the complete steps.
I understand that n^10/2^n is a null sequence based on standard null
sequences.
I couldn't convince myself why 2^n / n^10 is not a null sequence. I
understand that using Reciprocal Rule that this sequence tends to
infinity, but I do not want the argument to go that way. I want to see
a fresh set of arguments from a different angle.
.
- References:
- Solve This Sequences Problem
- From: fiazidris
- Re: Solve This Sequences Problem
- From: mensanator@xxxxxxxxxxx
- Solve This Sequences Problem
- Prev by Date: Product of multiple Normal Distributions
- Next by Date: Math as an Experimental Science
- Previous by thread: Re: Solve This Sequences Problem
- Next by thread: Re: Solve This Sequences Problem
- Index(es):
Relevant Pages
|
