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Math Help - prove convergence and find sum

  1. #1
    Eater of Worlds
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    prove convergence and find sum

    Let a_{n}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{(-1)^{n-1}}{n}-ln(2)

    Prove \sum_{n=1}^{\infty}a_{n} converges and find its sum.

    This shouldn't be too bad. Just a neat series.

    I started with:

    a_{n}=\int_{0}^{1}\left(1-x+x^{2}-....+(-1)^{n-1}x^{n-1}\right)-\int_{0}^{1}\frac{1}{x+1}dx
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by galactus View Post
    Let a_{n}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{(-1)^{n-1}}{n}-ln(2)

    Prove \sum_{n=1}^{\infty}a_{n} converges and find its sum.

    This shouldn't be too bad. Just a neat series.

    I started with:

    a_{n}=\int_{0}^{1}\left(1-x+x^{2}-....+(-1)^{n-1}x^{n-1}\right)-\int_{0}^{1}\frac{1}{x+1}dx
    1. The series converges by the Aternating series test.

    2. Abels theorem tells us that:

    If \sum_0^{\infty} a_n converges, then:

     <br />
\lim_{x \to 1^-}\sum_0^{\infty} a_n x^n = \sum_0^{\infty} a_n<br />

    Now:

    \ln(1+x)=\sum_1^{\infty} (-1)^{n-1}\frac{x^n}{n}

    for x \in (-1,1]

    Hence with just a bit of manipulation we conclude that:

    \left[\sum_1^{\infty}\frac{(-1)^{n-1}}{n}\right]-\ln(2)=0

    RonL
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by CaptainBlack View Post
    1. The series converges by the Aternating series test.

    2. Abels theorem tells us that:

    If \sum_0^{\infty} a_n converges, then:

     <br />
\lim_{x \to 1^-}\sum_0^{\infty} a_n x^n = \sum_0^{\infty} a_n<br />

    Now:

    \ln(1+x)=\sum_1^{\infty} (-1)^{n-1}\frac{x^n}{n}

    for x \in (-1,1]

    Hence with just a bit of manipulation we conclude that:

    \left[\sum_1^{\infty}\frac{(-1)^{n-1}}{n}\right]-\ln(2)=0
    Isn't the series that Galactus has to study \sum_{n=1}^{\infty} \left[-\ln2+\sum_{k=1}^n\frac{(-1)^{k-1}}{k} \right] ?
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  4. #4
    Eater of Worlds
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    Well done to all who attempted.

    Here is what I done.

    Starting where I left off:

    a_{n}=\int_{0}^{1}\left(1-x+x^{2}-x^{3}+.....+(-1)^{n-1}x^{n-1}\right)dx-\int_{0}^{1}\frac{1}{1+x}dx

    =\int_{0}^{1}\left(\frac{1+(-1)^{n-1}x^{n}}{1+x}\right)dx-\int_{0}^{1}\frac{1}{1+x}dx

    =\int_{0}^{1}\frac{(-1)^{n-1}x^{n}}{1+x}dx

    Now, for any integer N\geq 1, we get:

    \sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{N}\int_{0}^{1  }\frac{(-1)^{n-1}x^{n}}{1+x}dx

    =\int_{0}^{1}\frac{1}{1+x}\sum_{n=1}^{N}(-1)^{n-1}x^{n}dx

    =\int_{0}^{1}\frac{x+(-1)^{N+1}x^{N+1}}{(1+x)^{2}}dx

    =\int_{0}^{1}\frac{x}{(1+x)^{2}}dx+(-1)^{N+1}\int_{0}^{1}\frac{x^{N+1}}{(1+x)^{2}}dx

    Therefore,

    \left|\sum_{n=1}^{N}a_{n}-\int_{0}^{1}\frac{x}{(x+1)^{2}}dx\right|

    =\int_{0}^{1}\frac{x^{N+1}}{(1+x)^{2}}dx\leq \int_{0}^{1}x^{N+1}dx=\frac{1}{N+2}

    Now, if we let N\to {\infty} we can see that

    \sum_{n=1}^{\infty}a_{n} converges, and it has the sum

    \int_{0}^{1}\frac{x}{(1+x)^{2}}dx=\int_{0}^{1}\lef  t(\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\right)dx

    =\boxed{ln(2)-\frac{1}{2}}

    Which, it would appear, is pretty much the same idea that Paul had.
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  5. #5
    Global Moderator

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    See this.
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