January 11th, 2017, 11:14 AM  #1 
Newbie Joined: Jan 2017 From: Croatia Posts: 1 Thanks: 0  Little oh notation
First of all, I wanna say sorry for my bad english. Second, I just don't get the mentioned notation. Actually, I get how it works, but I don't get how many operands I have to have (you probably don't understand what I am talking about, so, I'll give the example $\displaystyle \boldsymbol{\lim _{n\to \infty }\left ( n\cdot (\frac{\sqrt[n]{e}+1}{\sqrt[n]{e}1}2\cdot n) \right )= }$ $\displaystyle =\lim _{n\to \infty }( n\cdot \frac{e^{\frac{1}{n}}+12n(e^{\frac{1}{n}}1)}{e^{\frac{1}{n}}1})$ $\displaystyle =\lim _{n\to \infty }( n\cdot \frac{1+\frac{1}{n}+\frac{1}{2n^{2}}+12n\cdot (1+ \frac{1}{n}+\frac{1}{2n^2}+\frac{1}{6n^3}+o(\frac{ 1}{n^3})1)}{\frac{1}{n}+\frac{1}{2n^2}+o(\frac{1}{n^2})})$ $\displaystyle =\lim _{n\to \infty }( n\cdot \frac{2+\frac{1}{n}+\frac{1}{2n^{2}}2\frac{1}{n}\frac{1}{3n^2}+o(\frac{1}{n^2})}{\frac{1}{n}+\frac {1}{2n^2}+o(\frac{1}{n^2})})$ $\displaystyle =\lim _{n\to \infty }( n\cdot \frac{\frac{1}{6}n+o(\frac{1}{n})}{\frac{1}{n}+ \frac{1}{2n^2}+o( \frac{1}{n^2})})$ $\displaystyle =\lim _{n\to \infty }(\frac{\frac{1}{6}+o(1)}{1+o(1)})$ $\displaystyle =\frac{1}{6}$ ...third row: there is sum of elements that are increasing their denominator's degree. How do I know how many elements I have to sum? Thank you. Last edited by Jovica; January 11th, 2017 at 11:18 AM. 
January 11th, 2017, 03:58 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,214 Thanks: 492 
You need to sum enough to get to the first surviving nonzero term.


Tags 
notation 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
SET Notation Help  Alexis87  Algebra  1  April 3rd, 2013 08:08 AM 
bigo notation  ploktoc  Number Theory  1  December 19th, 2012 05:49 PM 
Big O Notation  Ponchi182  Real Analysis  3  June 1st, 2011 12:48 PM 
notation help..  ElMarsh  Linear Algebra  9  November 5th, 2009 02:35 PM 
Big O notation  Christ1m  Applied Math  0  January 1st, 1970 12:00 AM 