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Math Help - Need help with summation/series.

  1. #1
    Junior Member
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    Need help with summation/series.




    (a) The series = ?



    b) The sequence = ?


    I have done the rest of the problems with out any problem.. but i cant seem to get this one right.

    Please help,
    Thank you
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by Diggidy View Post



    (a) The series = ?



    b) The sequence = ?


    I have done the rest of the problems with out any problem.. but i cant seem to get this one right.

    Please help,
    Thank you

    Need to think a bit about part a), but for now, here is some help with part b) if you set x = 2 / k (then note that x goes to zero as k goes to infinity.

    The typical term of the sequence becomes A_k = (1 + x) ^(2/x) =
    = [(1+x)^(1/x)]^2 whose limit, as x goes to zero, is e^2, where e is Euler's Constant. Some numerical investigation in Excel also verifies:

    k A_k

    1 3
    2 4
    3 4.62962963
    4 5.0625
    5 5.37824
    6 5.618655693
    7 5.8077951
    10 6.191736422
    15 6.536796047
    100 7.244646118
    1000 7.37431239
    10000 7.387578632

    Good luck!!
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    the series diverges as long as the limit of the sequence is finite.
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  4. #4
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    Hello, Diggidy!


    A_n \;=\;\left(1 + \frac{2}{n}\right)^n

    (b) The sequence \{A_n\}\:=\:?
    apcalculus is correct . . .

    Recall that: . \lim_{z\to\infty}\left(1+\frac{1}{z}\right)^z \:=\:e


    We have: . \left(1 + \frac{2}{n}\right)^n \;=\;\left(1 + \frac{1}{\frac{n}{2}}\right)^n  \;=\;\bigg[\left(1 + \frac{1}{\frac{n}{2}}\right)^n\bigg]^{\frac{2}{2}} \;= . \bigg[\left(1 + \frac{1}{\frac{n}{2}}\right)^{\frac{n}{2}}\bigg]^2


    If n\to\infty, then \tfrac{n}{2}\to\infty

    So we have: . \lim_{\frac{n}{2}\to\infty}\left[ \left(1 + \frac{1}{\frac{n}{2}}\right)^{\frac{n}{2}}\right]^2 \;= . \left[\lim_{\frac{n}{2}\to\infty}\left(1 + \frac{1}{\frac{n}{2}}\right)^{\frac{n}{2}}\right]^2 \;\;=\;\;e^2




    (a) The series: . \sum^{\infty}_{n=1}A_n \;=\;?

    Since the n^{th} term is does not go to zero, the series diverges.

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