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Math Help - More on series converge or diverge?

  1. #1
    Rui
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    More on series converge or diverge?

    Hi again, as I was doing more questions on this, I got stuck on two more questions.

    1. ∑(from n=1 to infinity) ((-1)^(n+1)/ √n )
    2. ∑(from n=1 to infinity) ( ln (1+1/n) )
    Well Q1, same as before I tried ratio test and it turned out weird...
    For Q2,not sure if I did it right, here's what I did. I compute the partial sums:
    s1= ln(1+1)
    s2= ln(1+1) + ln(1+1/2) > ln(1+1/2) + ln(1+1/2) = 2 ln(1+1/2)
    and so lim for n to infinity for ln(1+1/n) = infinity
    therefore the sequence of the partial sum diverges to infinity and so the sequence also diverges. Is this right?

    Thanks.
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by Rui View Post
    Hi again, as I was doing more questions on this, I got stuck on two more questions.

    1. ∑(from n=1 to infinity) ((-1)^(n+1)/ √n )
    You have to use the alternate series test, proving the absolute value of the summand is decreasing and that its limit when n tends to infinity is 0.
    Or you can take a look at the alternate Riemann series, which state that \sum \frac{(-1)^n}{n^a} converges if a>0
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by Rui View Post
    1. ∑(from n=1 to infinity) ((-1)^(n+1)/ √n )
    Use alternating series test:

    b_n = \frac 1{\sqrt{n}}

    b_{n+1} = \frac 1{\sqrt{n+1}}

    b_n \geq b_{n+1}

    \lim_{n\to\infty} b_n = \lim_{n\to\infty} \frac 1{\sqrt{n}} = 0

    So this satisfies the alternating series test, and is therefore convergent.
    Quote Originally Posted by Rui View Post
    2. ∑(from n=1 to infinity) ( ln (1+1/n) )
    let a_n = ln(1+1/n) ~~~~= ~~~~ln(\frac{n+1}n) ~~~~= ~~~~ln(n+1)-ln(n)

    Then lets evaluate this as a partial sum, looking at the telescoping aspect.

    \sum_{n=1}^k a_n = ln(2)-ln(1)~~~~+~~~~ln(3)-ln(2)~~~~+~~~~ln(4)-ln(3)~~~~+~~~~...~~~~+~~~~ln(k)-ln(k-1)

    cancel out common terms, and see that -ln(1) = 0 to get
    \sum_{n=1}^k a_n = ln(k)

    so \sum_{n=1}^\infty a_n = \lim_{k\to\infty} ln(k) which clearly diverges.
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