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Understanding Sequence of Partial Sum notation

  1. May 22, 2013 #1
    {Edit: as of 3:55 eastern time, made corrections to tex and itex mistakes}

    Is this all kosher in terms of demonstrating accuracy and comprehension of the notation [tex]{a_{1} + a_{2}...} = \lim_{n\rightarrow ∞ } \sum_{n=1}^{n} a_{n}[/tex]

    So the lower case represents sequences and upper case represents series.

    Sequence: [itex]a_{n} = a_{1}, a_{2}, a_{3}, a_{4}, a_{5}... [/itex]
    Series : [itex]A_{n} = a_{1} + a_{2} + a_{3} + a_{4} + a_{5}... [/itex]

    Sequence: [itex]b_{n} = a_{1}, (a_{1}+ a_{2}), (a_{1}+ a_{2}+ a_{3}), (a_{1}+ a_{2}+ a_{3}+ a_{4}), (a_{1}+ a_{2}+ a_{3}+ a_{4}+a_{5}), .... [/itex]
    Sequence: [itex]b_{n} = b_{1}, b_{2}, b_{3}, b_{4}, b_{5}... [/itex]
    Series : [itex]B_{n} = b_{1} + b_{2} + b_{3} + b_{4} + b_{5}... [/itex]

    So I can say, [itex]b_{5}=A_{5}[/itex]

    For concreteness, suppose a common ratio r = 1/2 is introduced, beginning at [itex]a_{1}=1[/itex],

    Sequence: [itex]a_{n} = 1, 1/2, 1/4, 1/8, 1/16...[/itex]
    Series: [itex]A_{n} = 1 + (1/2) + (1/4) + (1/8) + (1/16)...[/itex]

    Sequence: [itex]b_{n} = 1, (1 + 1/2), (1 + 1/2 + 1/4), (1 + 1/2 + 1/4 + 1/8), (1 + 1/2 + 1/4 + 1/8 + 1/16), .... [/itex]
    Sequence: [itex]b_{n} = 1, (3/2), (7/4), (15/8), (31/16)...[/itex]
    Series: [itex]B_{n} = 1 + (3/2) + (7/4) + (15/8) + (31/16)...[/itex]

    So [itex][b_{5}=A_{5}] = [31/16 = 1 + 1/2 + 1/4 + 1/8 + 1/16] = 1.9375[/itex]

    [itex]\lim_{n\rightarrow ∞ } A_{n} = \lim_{n\rightarrow ∞ } b_{n}[/itex] , and to be clear,

    [itex]\lim_{n\rightarrow ∞ }[/itex] of Series [itex]A_{n} = \lim_{n\rightarrow ∞ }[/itex] of Sequence [itex]b_{n}[/itex]

    And sigma notation [itex]\sum_{n=1}^{n} [/itex] are used to represent series, so

    [itex]A_{n}= \sum_{n=1}^{n} a_{n} = b_{n}[/itex]

    so for example, using subscript [itex]n=5[/itex]

    [itex]A_{5}= \sum_{n=1}^{5} a_{n} = b_{5} = 1.9375[/itex]

    And given that we are told the infinite series of [itex]A_{∞} = a_{} + a_{} +...a_{∞} = 2[/itex]

    Then we can write [itex]A_{∞}= \lim_{n\rightarrow ∞ } A_{n}=\lim_{n\rightarrow ∞ } b_{n}= \sum_{n=1}^{∞} a_{n} = \lim_{n\rightarrow ∞ } \sum_{n=1}^{n} a_{n}[/itex]

    And lastly, the series [itex]B_{n}[/itex] has no role in describing the infinite series [itex] a_{1}, a_{2}, a_{3}, a_{4}, a_{5}... [/itex]
    Last edited: May 22, 2013
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  3. May 22, 2013 #2


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    Notation etc. correct (except below). Why did you include bn and Bn in the discussion?

    Error in notation: You use n for both the running index and the upper limit. You need to use different letters, like k = (1,n).
  4. May 22, 2013 #3


    Staff: Mentor

    After a very quick scan, I didn't see anything glaringly wrong, but I'd like to give an example that puts all of the terms in context.

    Every infinite series is associated with two sequences: the sequence that makes up the terms in the infinite series, and the sequence of partial sums. We investigate the sequence of partial sums to determine whether the series converges.

    Here's the example - the series
    $$ \sum_{n = 0}^{\infty} \frac{1}{2^n}$$

    Expanded, this series looks like 1 + 1/2 + 1/4 + 1/8 + ... + 1/2n + ...
    The sequence of terms: {1, 1/2, 1/4, ..., 1/2n, ...}
    The sequence of partial sums: {1, 3/2, 7/4, 15/8, ...}
    $$ S_m= \sum_{n = 0}^m \frac{1}{2^n}$$

    It turns out that the sequence of partial sums converges (to 2), which means that the series converges to this same number. This is a very simple series (a geometric series), so I won't go into the details of why it converges.
  5. May 22, 2013 #4
    I used bn and Bn to represent a different set of terms that are being sequenced and summed.

    The an and An represent the sequence and series of {1, (1/2), (1/4)...} while bn and Bn represent the sequence and series of {1, (3/2), (7/4),...}

    I figure its the same concept as assigning f(x) to one function and g(x) to an entirely different function.

    Thanks, so I should use a different number, like k, in the running index like this:

    [tex]A_{n} = {a_{1} + a_{2} + a_{n}} = \sum_{k=1}^{n} a_{n}[/tex]

    or in the upper limit like this:

    [tex]A_{n} = {a_{1} + a_{2} + a_{n}} = \sum_{n=1}^{k} a_{n}[/tex]

    I would guess the latter as Mark44 demonstrates: $$ S_m= \sum_{n = 0}^m \frac{1}{2^n}$$

    Then that means that I'd have to change [itex]A_{n} to A_{k}[/itex] so that:

    [tex]A_{k} = {a_{1} + a_{2} + a_{k}} = \sum_{n=1}^{k} a_{n}[/tex]

    However, it is not completely obvious to me why the n can't be the only variable. Why the need to introduce k? Could it be because k represents a finite number while n represents a variable of a function/term?

    Mark44, thanks for the very matter of fact clarification. Every infinite series is associated with two sequences, sequence of terms and sequence of partial sums. Btw, Mark44, in the paragraph above, is the reasoning I've suggested for why n should be replaced with k correct?
    Last edited: May 22, 2013
  6. May 22, 2013 #5


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    An = a1 + a2 + ... + an
    As you wrote it, your series has just three terms. The ellipsis (...) implies that the terms in the series continue in a certain pattern.
    $$=\sum_{k=1}^{n} a_{k}$$
    In the summation, the index variable is k, which takes on the values 1, 2, 3, ..., n.
    This is wrong, too. One variable represents the number of terms in your partial sum (n). The index variable is used to identify each of the terms in the sum. The sum on the left runs from index 1 to index n, so the summation has to do the same. The index variable can't be n, but it could by i or j or k or whatever.
    Again, you need a different variable to represent the various index values and one to represent the highest index.
  7. May 22, 2013 #6
    Thanks Mark44, would the notation be like this? EDIT: In other words, can you point out where the variables may be contradicting.

    for finite series

    [tex] [\frac{1}{2^{k}} + \frac{1}{2^{k}} + \frac{1}{2^{k}} + ... \frac{1}{2^{n}}] \Rightarrow [a_{1} + a_{2} + a_{3} +,... a_{n}] = A_{n} = \sum_{k=0}^{n} a_{k} = \sum_{k=0}^{n} \frac{1}{2^{k}} = \lim_{k\rightarrow n } A_{k} [/tex]

    for infinite series

    [tex] [{\frac{1}{2^{k}} + \frac{1}{2^{k}} + \frac{1}{2^{k}} + ...\frac{1}{2^{∞}}}] \Rightarrow [{a_{1} + a_{2} + a_{3} +, ... a_{∞}}] = A_{∞} = \sum_{k=1}^{∞} a_{k} = \sum_{k=0}^{∞} \frac{1}{2^{k}} = \lim_{k\rightarrow ∞ } A_{k} = \lim_{n\rightarrow ∞ } \sum_{k=1}^{n} a_{k} [/tex]

    I get the feeling the variable k can not be repeated over and over like I've written. Perhaps the series should be written this way [tex] [{\frac{1}{2^{k}} + \frac{1}{2^{k+1}} + \frac{1}{2^{k+2}} + ...\frac{1}{2^{n}}}][/tex]
    Last edited: May 22, 2013
  8. May 23, 2013 #7
    After thinking about it more indepthly, I think I'm more confused now than before posting the original questions.

    taking it a step back to sequences,

    [itex]a_{n} = a_{1}(\frac{1}{2})^{n-n}, a_{1}(\frac{1}{2})^{1}, a_{1}(\frac{1}{2})^{2}... a_{1}(\frac{1}{2})^{n-1} = a_{1}, a_{2}, a_{3}...a_{n}[/itex]

    [itex]a_{1}[/itex] corresponds to [itex]a_{1}(\frac{1}{2})^{n-n}[/itex]
    [itex]a_{2}[/itex] corresponds to [itex]a_{1}(\frac{1}{2})^{1}[/itex]
    [itex]a_{n}[/itex] corresponds to [itex]a_{1}(\frac{1}{2})^{n-1}[/itex]

    Then for series,

    [itex]A_{n} = (1)(\frac{1}{2})^{0} + (1)(\frac{1}{2})^{1} + (1)(\frac{1}{2})^{2} +... (1)(\frac{1}{2})^{n-1}[/itex]

    However, when writing a series, shouldn't the subscript n on the left side match exactly the last term's variable n. That is, shouldn't the last term somehow be represented like this:

    [itex]A_{n} = (1)(\frac{1}{2})^{0} + (1)(\frac{1}{2})^{1} + (1)(\frac{1}{2})^{2} +... (1)(\frac{1}{2})^{n}[/itex]

    Or is that a false assumption? Does it only need to match the general term numbers, as in the last term of the far right side of the equation:

    [itex]A_{n} = (1)(\frac{1}{2})^{0} + (1)(\frac{1}{2})^{1} + (1)(\frac{1}{2})^{2} +... (1)(\frac{1}{2})^{n-1} = a_{1} + a_{2}+ a_{3} +...a_{n}[/itex]

    [[EDIT: changed the variable to match the index in the summation notation below at 9:30 am]]

    If this is the case, then the sigma notation has two different indexes depending on whether the function or the general term are being represented. That is:

    [itex]\sum_{i=1}^{n} a_{i}[/itex]

    [itex]\sum_{i=0}^{n} (1)(\frac{1}{2})^{i}[/itex]

    And I'm still sketchy on why the index must be a different variable. Does it mean this:

    [itex]\sum_{i=1}^{n} a_{i} = a_{i}......a_{n}[/itex]

    [itex]\sum_{i=0}^{n} (1)(\frac{1}{2})^{i} = a_{1}r^{i}.....a_{1}r^{n}[/itex]
    Last edited: May 23, 2013
  9. May 23, 2013 #8
    I'll get it with some practice and further thought. Thanks for clearing it up for me, the rest is up to me to think about. Thanks.
  10. May 24, 2013 #9


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    Last 2 lines should read:
    [itex]\sum_{i=1}^{n} a_{i} = a_{1}+.....+a_{n}[/itex]

    [itex]\sum_{i=0}^{n} (1)(\frac{1}{2})^{i} = a_{1}r^{0}+.....+a_{1}r^{n}[/itex]

    Also, for the last line, you know r = 1/2 and a1 = 1, so why use the symbols?
  11. May 29, 2013 #10
    Thanks. Using the symbols just to get an idea of how to write the various statements in general terms. I was confusing how the index and upper limit variables in the sigma notation correspond with writing out the series in general terms. I've got the idea better now. I've since written it down on paper using lowercase n for the index and capital N for the upper bound. Appreciate all the help in this thread.
  12. May 29, 2013 #11


    Staff: Mentor

    With corrections and comments:
    $$ A_n = a_1 + a_2 + a_3 + ... + a_n = \frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + ... \frac{1}{2^{n}} = \sum_{k=1}^{n} \frac{1}{2^{k}} $$
    1. The first part of what you wrote above doesn't make any sense. There are three terms of 1/2k, followed by an ellipsis (...), but there isn't a perceivable pattern.
    2. For consistency, all of the partial sums start with index 1. If you want to start them at index 0, the first term of each sum should be a0, which is 1/20.

    Same comments as above. In addition, we never write expressions such as 1/2. To denote an infinite series, write it as a1 + a2 + ... + an + ... The final ellipsis denotes that the series continues on indefinitely, which is something a finite series doesn't do.
    No. See above.
  13. May 29, 2013 #12
    Is this a more accurate way to express finite series in general terms.

    for finite series

    [tex] A_{N} = \sum_{n=0}^{N} a_{n} = \sum_{n=0}^{N} \frac{1}{2^{n}} = \lim_{n\rightarrow N } A_{n} [/tex]

    [tex] A_{N} = \sum_{n=0}^{N} \frac{1}{2^{n}} = [{\frac{1}{2^{0}} + \frac{1}{2^{1}} + \frac{1}{2^{2}} + ...\frac{1}{2^{N - 1}} + \frac{1}{2^{N}}}][/tex]
  14. Jul 28, 2013 #13
    My apologies, this is the first week I'm able to revisit math in about a couple months. Feels like I forget more than I learn with all these gaps in time spent away from math study due to my current job. Hopefully I can pick up where I left off. Question I had in this thread was:

    I wrote:

    Your correction:

    I followed up:

    Your replied:

    I am confused by your reply here because it seems to me that the error you highlighted was corrected in the follow up.

    Would this not be correct? If not correct, is it because the way I wrote this implies that the number replacing the k variable in each consecutive term stays the same when in fact the number replacing the k variable in each consecutive term is supposed to increase by 1 consecutively?

    [tex]A_{n} = a_{0} + a_{1} + a_{2} +... a_{n} = \sum_{k=0}^{n} \frac{1}{2^{k}} = [{\frac{1}{2^{k}} + \frac{1}{2^{k+1}} + \frac{1}{2^{k+2}} + ...\frac{1}{2^{n}}}] = [{\frac{1}{2^{0}} + \frac{1}{2^{0+1}} + \frac{1}{2^{0+2}} + ...\frac{1}{2^{n}}}] [/tex] [tex] = [{\frac{1}{1} + \frac{1}{2^{1}} + \frac{1}{2^{2}} + ...\frac{1}{2^{n}}}] = [{1 + \frac{1}{2} + \frac{1}{4} + ...\frac{1}{2^{n}}}] [/tex]

    In other words, if I could attempt to answer my own question, would this possibly be a correct way of expressing the series, if the above is incorrect?

    [tex]A_{n} = a_{0} + a_{1} + a_{2} +... a_{n} = \sum_{k=0}^{n} \frac{1}{2^{k}} = [{\frac{1}{2^{0}} + \frac{1}{2^{1}} + \frac{1}{2^{2}} + ...\frac{1}{2^{n}}}] = [{1 + \frac{1}{2} + \frac{1}{4} + ...\frac{1}{2^{n}}}] [/tex]

  15. Jul 29, 2013 #14


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    The above is mostly correct, except for this part.
    $$\sum_{k=0}^{n} \frac{1}{2^{k}} = [{\frac{1}{2^{k}} + \frac{1}{2^{k+1}} + \frac{1}{2^{k+2}} + ...\frac{1}{2^{n}}}] $$

    This is what it should say:
    $$\sum_{k=0}^{n} \frac{1}{2^{k}} = \frac{1}{2^{0}} + \frac{1}{2^{1}} + \frac{1}{2^{2}} + ... +\frac{1}{2^{n}} $$

    To expand the summation notation, the index variable k is replaced by 0, 1, 2, ... , n and the n + 1 terms are added together. In the expanded form, k should NOT appear.

    One minor point: When you write ..., put a plus sign in front of and after the dots.
    The above looks fine, although you should write + ... + in both places. Again, this is a minor point, and most people would understand what you mean.
  16. Jul 29, 2013 #15
    Thank you Mark.

    In the following statement:

    What do you mean by:

    "...n and the n + 1 terms are added together."

    I did not write n + 1 anywhere and am wondering what you have meant by it.
  17. Jul 29, 2013 #16


    Staff: Mentor

    It might have helped if I had added one more comma, like so: To expand the summation notation, the index variable k is replaced by 0, 1, 2, ... , n, and the n + 1 terms are added together.

    The index variable k takes on these values: 0, 1, 2, ..., n.
    By inspection, k takes on n + 1 values. For example, if the list of values were 0, 1, 2, 3, 4, and 5, that would be 5 + 1 (=6) values.

    For each value of k there is an expression that gets added in the summation. For what you wrote, the partial sum consists of n + 1 terms.
  18. Aug 5, 2013 #17
    Thank you Mark,

    These statements should be correct then:

    This has n+1 terms. That is, k takes on n+1 values in this statement:
    [tex]\sum_{k=0}^{n} \frac{1}{2^{k}} = [{\frac{1}{2^{0}} + \frac{1}{2^{1}} + \frac{1}{2^{2}} + ... +\frac{1}{2^{n}}}] = [{1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^{n}}}] [/tex]

    While this has n terms. That is, k takes on n values in this statement:
    [tex]\sum_{k=1}^{n} \frac{1}{2^{k}} = [{\frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + ... +\frac{1}{2^{n}}}] = [{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{2^{n}}}] [/tex]
  19. Aug 5, 2013 #18


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