January 11th, 2017, 10:14 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 10:18 AM. 
January 11th, 2017, 02:58 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,540 Thanks: 591 
You need to sum enough to get to the first surviving nonzero term.


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