My Math Forum Little oh notation

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 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 $\displaystyle \boldsymbol{\lim _{n\to \infty }\left ( n\cdot (\frac{\sqrt[n]{e}+1}{\sqrt[n]{e}-1}-2\cdot n) \right )= }$ $\displaystyle =\lim _{n\to \infty }( n\cdot \frac{e^{\frac{1}{n}}+1-2n(e^{\frac{1}{n}}-1)}{e^{\frac{1}{n}}-1})$ $\displaystyle =\lim _{n\to \infty }( n\cdot \frac{1+\frac{1}{n}+\frac{1}{2n^{2}}+1-2n\cdot (1+ \frac{1}{n}+\frac{1}{2n^2}+\frac{1}{6n^3}+o(\frac{ 1}{n^3})-1)}{\frac{1}{n}+\frac{1}{2n^2}+o(\frac{1}{n^2})})$ $\displaystyle =\lim _{n\to \infty }( n\cdot \frac{2+\frac{1}{n}+\frac{1}{2n^{2}}-2-\frac{1}{n}-\frac{1}{3n^2}+o(\frac{1}{n^2})}{\frac{1}{n}+\frac {1}{2n^2}+o(\frac{1}{n^2})})$ $\displaystyle =\lim _{n\to \infty }( n\cdot \frac{\frac{1}{6}n+o(\frac{1}{n})}{\frac{1}{n}+ \frac{1}{2n^2}+o( \frac{1}{n^2})})$ $\displaystyle =\lim _{n\to \infty }(\frac{\frac{1}{6}+o(1)}{1+o(1)})$ $\displaystyle =\frac{1}{6}$ ...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 non-zero term.

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