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Infinite sequences and series help!

  1. Dec 7, 2012 #1
    Hi I don't understand the logic in the picture i added.

    They say that "that sum of the series = the limit of the sequence"

    The limit is 2/3 BUT the sum, Ʃ, must be 2*1/(3*1+5) + (2*2/(2*3+5) + 2*3/(2*3+5) ...+
    Which is obviously much larger than 2/3 if all the terms are added together?? it's like adding 2/3+2/3+2/3 which is = 6/2 =2??

    Is there something I'm misunderstanding here?

    Attached Files:

  2. jcsd
  3. Dec 7, 2012 #2


    Staff: Mentor

    what is the a sub n term? the a sub n terms are added together to get the s sub n

    the sequence referred to is the s sub n terms as defined by the 2*n/(3*n+5) you don't sum these. Instead you must evaluate the term as n approaches infinity.

    so that when n is very large the sequence terms become 2*n / 3*n and the n drops out to leave 2/3
  4. Dec 7, 2012 #3
    Well if we have the series Ʃan= 1/2n i can see that the sum would add up to one.

    Ahhh So the limit of one Lim 1 = 1 = the sum?? Is that how I should understand it?

    Attached Files:

    • sum.JPG
      File size:
      15.6 KB
  5. Dec 7, 2012 #4
    Ahh so the partial sequence of Sn adds up to 2n/(3n+5) as you said (just like 1/2^2 adds up to one). so if we add up all the terms of some imaginary sequence we would get the sum to be 2n/(3n+5), so the sum is the end result and NOT 2*1/(3*1+5) + (2*2/(2*3+5) + 2*3/(2*3+5) ...+ which is the result of adding up the sums.

    So the limit is the the sum because the sum is 2n/(3n+5) and when n gets big it's 2/3. Is that correctly understood? And thank you :)
  6. Dec 7, 2012 #5
    So is it correctly understood that the limit of a sequence 1/2^n is zero but the limit of the partial sums of that sequence is 1? That's the diffrence?
  7. Dec 7, 2012 #6


    Staff: Mentor

    Don't say adds up, its not adding up its a sequence that approaches a limit, the limit is 2/3

    Look at a simpler example:

    Ʃ(2n+1) = n^2

    an = (2n+1)

    sn = n^2

    so the sequence of sn is { 1, 4, 9, 16, 25 ... }

    so as n approaches infinity what is the last term of the sequence of sn
  8. Dec 7, 2012 #7
    don't look at it that way

    divide each term by the highest power variable and take each portion of the equations limit, separately.

    a quick way of doing it is to take a ratio of coefficients of variables in this case of an
  9. Dec 7, 2012 #8
    Hmm well Okay first, Ʃ(2n+1) means 2*1+1 +2*2+1 ...(2n+1) this is the action of adding up a sequence of numbers, right? Isn't that why we use a summation symbol? Isn't it the definition of summation "Sum" = add up?

    Now follow me here (and please tell me it's correctly understood :-) then I'll be happy )
    The sum of this series 1/2+1/4+1/8...+ has a pattern 1/2^n and we denote it Ʃ(1/2^n ), By taking a partial sum Sn= 1/2+1/4+1/8 We check if this sum approaches a limit (in this case Lim Sn=1)
    So the sum of the series Ʃ(1/2^n ) is the limit of the sequence of partial sums, right??

    The limit of 1/2^n approaches zero but the limit of the partial sum Sn approaches one.

    This must be correct, please :D
    Last edited: Dec 7, 2012
  10. Dec 7, 2012 #9


    Staff: Mentor

    With respect to your prior statement, it would be better to say:

    partial sum of Sn is given by 2n/(3n+5) and this approaches 2/3 as n approaches infinity.

    perhaps we're splitting hairs here but I didn't want you to think that the 2n/(3n+5)

    is the same as s1+s2+s3...

    With respect to this statement it appears to be correct.
  11. Dec 7, 2012 #10
    Thank you! :)

    I see because you wanted me to understand that 2n/(3n+5) is the infinite series (when n goes to infinity) and s1+s2+s3... is a sequence of numbers.
  12. Dec 7, 2012 #11


    Staff: Mentor

    The infinite series is
    $$ \sum_{n = 1}^{\infty} \frac{2n}{3n + 5}$$

    2n/(3n + 5) is merely the n-th term in the series or the n-th term in the sequence that underlies the series.
    No, s1 + s2 + s3 +... is a sum (or infinite series). The ellipsis (...) indicates that this sum continues indefinitely, following the same pattern.

    Every infinite series involves two sequences:
    1. The sequence that makes up the terms in the series.
    2. The sequence of partial sums.

    Using the first example above, the sequence of terms would be
    {2/8, 4/11, 6/14, ..., 2n/(3n + 5), ...}
    The seqence of partial sums would be
    {2/8, 2/8 + 4/11, 2/8 + 4/11 + 6/14, ...}

    (I haven't bothered to simplify the fractions or combine them.)

    A series converges if its sequence of partial sums converges.
  13. Dec 7, 2012 #12
    Thank you I finally understand Sn now completely,
    S1 = a1
    S2= a1 +a1
    S3 = a1+a2+a3.

    So the n-th term of an is 2n/(3n + 5) and the n-th term of Sn as i goes from 1 to n gives approaches the limit/sum for the finite series if it converges.
  14. Dec 7, 2012 #13
    But wait, this part: 2n/(3n + 5) is merely the n-th term in the series or the n-th term in the sequence that underlies the series.

    What's the difference between the series and the sequence that underlies the series?
    The series is the infinite sequence of numbers right? and the sequence is a part of that series?
  15. Dec 7, 2012 #14
    And to this "1. The sequence that makes up the terms in the series."
    A term is 2/8 , so don't you mean "the terms that make up the sequence (a1 +a2+a3) in the series"
  16. Dec 7, 2012 #15


    Staff: Mentor

    The series is what you get when you add all of the terms in the sequence.
    The sequence is a list, albeit an infinitely long list: {a1, a2, a3 ,,, an...}
    The series (AKA infinite series) is the sum that you get when you add all of the terms in the sequence.

    I think you are confusing the sequence that represents the terms, which is abbreviated as {an}, with the sequence of partial sums. {S1, S2, S3, ... , Sn, ...}

    Here S1 = a1
    S2 = a1 + a2 = S1 + a2
    S3 = a1 + a2 + a3= S2 + a3
    and so on.

    In general, Sn = ## \sum_{i = 1}^n a_n##, which is a finite sum.
  17. Dec 7, 2012 #16
    I'm sory, You are right I'm confusing things together. I sat down, re-read everything, and here it is. I understand it now, I really spent time thinking about it.

    We have a sequence of number fx the sequence 1/2n, and we can test the limit of such a sequence, by doing this we want to find out what the sequence approaches as n → ∞
    Lim1/2n = 0
    If we add the terms of an infinite sequence we get what we call an Infinite series. The series is a number/sum denoted Ʃ(an) consisting of the addition of the numbers in the sequence an.

    Instead of adding the whole sequence to infinity, we add a part of the sequence. We call this The partial Sums, denoted Sn.
    Sn = A1+a2+a3…an(…=infinity)
    We can test if this series of numbers converges toward a finite number, or if it diverges to infinity by using partial sum.

    If we have a sequence an = 1/2n
    The sequence is: 1/2 ,1/4, 1/8…,1/nn...(...= going to infinity)

    The limit of the sequence if we let n go toward infinity is Lim 1/2n =0

    On the other hand The series Ʃ(1/2n) also has a limit. We can prove this by a rule that says if the Sequece of the partial sum {Sn} is convergent (which it is, because the limit of the sequence {Sn}=1/2 ,1/2+1/4,1/2+1/4+ 1/8…,1/nn is 1} then the series Ʃ(1/2n) is convergent, and it is because it converges toward 1.
    Last edited: Dec 7, 2012
  18. Dec 7, 2012 #17
    You were right, I was confusing the sequence of terms with the sequence of partial sums, that was what confused me so much in my calc book. Thank you! :)
  19. Dec 7, 2012 #18


    Staff: Mentor

    No need for apology. This is complicated stuff when you're first exposed to it.
    I can tell. It looks like you understand pretty well.

    There are a couple of nits below, but otherwise you seem to have a good handle on the ideas.
    If the series is convergent, then it represents a number. Other series are divergent, so they don't represent a number.
    Sn represents the sum of a finite number of terms for each value of n.
    Sn = a1 + a2 + ... + an. However, as n increases, the number of terms that make up Sn gets larger, but there are always n terms.
    The sequence of partial sums doesn't have to grow unboundedly large for the series to diverge. For example, in this series -- ##\sum_{i = 1}^{\infty} (-1)^n## -- the sequence of partial sums is {-1, 0, -1, 0, -1, 0, ...}. Notice that I am actually adding the terms in the series to get this sequence of partial sums: {-1, -1 + 1, -1 + 1 - 1, ...}. This sequence (of partial sums) is not growing larger, but the terms in the sequence never settle in to a particular value. Hence ##\sum_{i = 1}^{\infty} (-1)^n## is divergent.
    The last term that you show should be 1/2n.
    The last term that you show above should be 1/2 + 1/4 + 1/8 + ... + 1/2n.

    A little work shows that the sequence of partial sums can be written as {1/2, 3/4, 7/8, ... , (2n - 1)/22, ...}
    We say "converges to 1", not "toward".

    Good work!
  20. Dec 7, 2012 #19


    Staff: Mentor

    Great post Mark44!
  21. Dec 7, 2012 #20
    Man, I really appreciate all your help. You are great at explaining this topic!

    So If if the sequence of a partial sum is (2n - 1)/22, Then this must mean that there is a pattern, just like for a sequence an and for a series Ʃan. So if the partial sum Sn = (2n - 1)/22 what exactly does that tell us?

    It must be the partial sum of the n-th term in the sequence an,

    I remember from the text "The sum of a series is the limit of the sequence of partial sums", so the sum of the series Ʃ1/2n must be the limit of Sn? But does it make sense to speak about a limit of Sn = (2n - 1)/22 which is a finite number? It's like saying "the limit of 25 = 25" right?
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