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Question
A sequence of real number (a_n)_n=1(n goes to infinity) converges to a real number l if for every ε>0 there is an n_0 such that |a_n-l| <ε forall n>n_0
Sir please describe the avobe defination
Let's explain it by an example : Suppose we want to chick the convergence of the sequence
a(n)=1/n , when n goes to INF . We already know that as n goes to INF , the sequence 1/n
converge to zero !
The definition implies that : If we are interested in an accuracy of a very small ε , then there must be a very big number no , that from this number on , we will certainly achieve the accuracy .
Let's say we want to know from which number on the sequence a(n)=1/n converge to 0 in an
accuracy of 0.001 ? That means |1/n - 0|<0.001 or 1/n<0.001 . It's pretty obvious that
no=1/0.001=1000 . When n reaches 1000 %26 beyond, then certainly a(n)<0.001 .
So, the conclusion will be , for every ε , no=1/ε .
Further more , if a(n)=1/n^2 then no=1/ε^2 .
Alon.
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Answers by Expert:
Alon Mandes
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Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
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1. I'm a team member of mathnerds (math site for answering questions)
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5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" ,
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M.A in Mathematics & Bs.c in Electronics.
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