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Can someone check my explanation of sequences and series?

  1. Apr 12, 2014 #1
    It's been a while since I've dealt with sequences and series. Here is my explanation of sequences and series and let me know if I am right or wrong.

    A sequence is just a list of numbers. By convention, we use the letter ##a## for sequences and they are written in a form like so: ##a_1,a_2,a_3,a_4,...##

    A sequence can be finite or infinite.

    1,2,3,4 is a finite sequence. 1,2,3,4,... is an infinite sequence.

    An arithmetic sequence, for some constant d: ##a_n=a_0+dn##
    A geometric sequence, for some constant r: ##a_{ n }=a_{ 0 }r^{ n }##

    A series is the sum of the terms of a sequence. By convention, is there a letter for series? I can't remember. Let us use the letter S in the meantime. Series are written like so: ##S_1,S_2,S_3,S_4,...##

    Let Sn be the series of the finite sequence mentioned earlier. S1=1. S2=3. S3=6. S4=10.

    Let Sn be the series of the infinite sequence mentioned earlier. S=∞

    I feel like I'm doing something wrong. Can anyone briefly mention which parts are wrong?
  2. jcsd
  3. Apr 12, 2014 #2


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    Except for minor typo (you have Sn when you meant S in last line), looks fine.
  4. Apr 13, 2014 #3
    When you explain arithmetic and geometric sequences, the general term expression(s) are given by:
    [itex] a_n = a_0 + d(n-1) [/itex]
    [itex] g_n = g_0 \cdot r^{n-1} [/itex]
    The [itex] n-1 [/itex] is necessary to generate the term; for example, the sequence
    [itex] 1, 2, 3, 4, 5,... [/itex]
    is arithmetic, and if I wanted to generate the 6th term of the sequence, I would use:
    [itex] a_6 = 1 + 1(6-1) = 6[/itex]
    Had I used your expression, [itex] a_n = a_0 + dn [/itex], then the term would be:
    [itex] a_6 = 1 + 1(6) = 7[/itex], which is obviously not the 6th term of the sequence.
    The same goes for geometric sequences. Consider
    [itex] 2, 4, 8, 16, 32,...[/itex]
    If I wanted to generate the 6th term of this sequence, I would use:
    [itex] g_6 = 2 \cdot (2)^{6-1} = 2 \cdot 32 = 64 [/itex]
    Again, if I used your expression:
    [itex] g_6 = 2 \cdot (2)^{6} = 2 \cdot 64 = 128 [/itex]
    So really what your expressions do is generate the [itex] n + 1 ^{th} [/itex] term of the sequence, rather than the [itex] n^{th} [/itex]
    Your explanations of series are also fine, but I might add the more formal definition of a series:
    [itex] S_n = \sum_{i=0}^{n} t_i [/itex] which is really just the sum of some number of terms.Furthermore, an infinite sequence does not always add to infinity, as a geometric sequence with a ratio [itex] r [/itex] such that [itex]-1 < r < 1 [/itex] converges as [itex] n \rightarrow \infty [/itex]. The example that you gave, though, will diverge to [itex] \infty [/itex].
  5. Apr 14, 2014 #4


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    Well, this would imply that [itex] a_0 = a_0 -d [/itex], which is not quite right. You probably mean [itex]a_{n-1}[/itex], but there's nothing wrong with Visheras explanation.
    Last edited: Apr 14, 2014
  6. Apr 14, 2014 #5
    I'm not sure what you're arguing, as [itex] a_{n+1} [/itex] implies a recursion, in which case (for an arithmetic progression) [itex] a_{n+1} = a_n + d [/itex]
    [itex] a_0 = a_0
    a_1 = a_0 + d
    a_2 = a_1 + d = (a_0 + d) + d = a_0 + 2d
    a_3 = a_2 + d = (a_0 + 2d) + d = a_0 + 3d
    a_4 = a_3 + d = (a_0 + 3d) + d = a_0 + 4d
    a_n = a_{n-1} + d = (a_0 + (n-1)d) \rightarrow a_n = a_0 + d(n-1)
    Furthermore, for a geometric sequence:
    [itex] g_0 = g_0
    g_1 = g_0 \cdot r
    g_2 = g_1 \cdot r = (g_0 \cdot r) \cdot r = g_0 \cdot r^2
    g_3 = g_2 \cdot r = (g_0 \cdot r^2) \cdot r = g_0 \cdot r^3
    g_n = g_0 \cdot r^{n-1}
    Keep in mind that [itex] a_n [/itex] or [itex] g_n [/itex] is the [itex] n + 1^{th} [/itex] term.
  7. Apr 14, 2014 #6


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    If you look at the sentence I quoted you will see what I mean.
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