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Definition of sequence and series

  1. Sep 26, 2011 #1
    Hi!
    While studying sequence and series, I've gotten some misunderstandings in the definitions of sequence and series.
    What I know about the definitions of sequence and series is as follows below
    ; a sequence of field of real numbers is defined as a function mapping of the set of all positive integers into the field , and a series of real field is defined to be a sequence whose range consists of partial sums [itex]\sum[/itex] [itex]^{k}_{n=1}[/itex]a[itex]_{n}[/itex].
    But, in some other books, the definitions of them are a little different from what i've written as above. I mean that in some books, they give a function from { k[itex]\in[/itex] Z : k[itex]\geq[/itex]k[itex]_{0}[/itex] for an integer k[itex]_{0}[/itex]} into the real field as a definition of a sequence.
    What i'd like to ask you in this situation is whether there're no problems even if I use the different definitions of a sequence when I prove all theorems relating with the sequence ;
    for example, Bolzano-weierstrass theorem, additivity of limit of two sequences, etc.

    Furthermore, when we test the convergence of a series of some sepecial kinds, as you know, we usually use root, ratio tests. At first, I've learned that this tests are used when the sequence of a series runs from 1 to infinity. But, I recently found in some books that those methods are also used in determining the convergence of series running from some integer k[itex]_{0}[/itex] into infinity. That means the sequence of the series starts from the integer k[itex]_{0}[/itex].I wanna ask you if the root or ratio test can be used even in dealing with the series whose sequence runs from a integer k[itex]_{0}[/itex]. If so, I also ask you the rigorous definition of series of this kind [itex]\sum[/itex][itex]^{infinity}_{n=k_{0}}[/itex]a[itex]_{n}[/itex].

    Thank you for reading my long questions.
    I'll wait for your answers!!!
    God bless you! Have a nice day!
     
    Last edited: Sep 26, 2011
  2. jcsd
  3. Sep 26, 2011 #2

    CompuChip

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    Whether you start your sequence at 0, 1 or any other positive integer k0 does not matter. It is an easy exercise to prove that they are equivalent (basically finding a bijection is enough - and the bijection is just shifting the index by k0).

    You can define an "infinite" series as the limit of the partial sums:
    [tex]\sum_{n = 0}^\infty a_n = \lim_{N \to \infty} \sum_{n = 0}^N a_n[/tex]
     
  4. Sep 26, 2011 #3
    Do you mean that I can reach to the same theorem (but the definition of sequence in the theorem is a little different) concerned with a sequence even though I use different definitions of a sequence ??

    In the above equation, does the symbol N run from 0? or 1?
     
  5. Sep 26, 2011 #4

    CompuChip

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    Yes, because I claim that the definition is equivalent for all practical purposes as they are related by a simple shift.

    I don't really understand the second question... N is the limit variable.
     
  6. Sep 26, 2011 #5
    What I know about the definition of limit of a given sequence <b[itex]_{N}[/itex]>( it starts from 1;b[itex]_{1}[/itex],b[itex]_{2}[/itex], ...) is as follows; the sequence <b[itex]_{N}[/itex]> converges to b[itex]_{0}[/itex] if for any [itex]\epsilon[/itex]>0, [itex]\exists[/itex]M>0 such that N>M implies [itex]\left|[/itex]b[itex]_{N}[/itex]-b[itex]_{0}[/itex][itex]\right|[/itex]<[itex]\epsilon[/itex].
    In the above equation you show me, it's a limit of a sequence <b[itex]_{N}[/itex]>=[itex]\sum[/itex][itex]^{N}_{n=0}[/itex]a[itex]_{n}[/itex]. If the sequence <b[itex]_{N}[/itex]> starts from 1, then I can apply the definition of limit of a sequence to this sequence. But if the sequence <b[itex]_{N}[/itex]> starts from 0, then it's confusing for me to apply the definition of limit of sequence as I mentioned above. But You told me that it doesn't matter whether I use different definitions of a sequence. So I wonder if in the case of the sequence <b[itex]_{N}[/itex]> there's another definition of limit of sequence, too. ;;Because of this I'm confused .
     
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