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Evaluate sum(n=2,n=infinity) ln(1-1/n^2) if it exists.

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    1-1/n^2 = (n^2 - 1)/n^2 =(n+1)(n-1)/n^2

    ------>

    Log(1-1/n^2) =

    Log(n+1) + Log(n-1) - 2 Log(n).


    If we define the difference operator Delta as:

    (Delta f)(n) = f(n+1) - f(n)

    Then applying Delta twice gives:

    (Delta^2 f)(n) =

    f(n+2) - 2f(n+1) + f(n)

    The formula for Log(1-1/n^2) has the same structure. We see that:

    Log(1-1/n^2) = Delta^2 f(n)

    with:

    f(n) = log(n-1)

    In general, we have:

    Sum from n = r to s of (Delta g)(n) =

    g(r+1) - g(r) + g(r+2) - g(r+1) + ...
    + g(s) - g(s-1) + g(s+1) - g(s) =

    g(s+1) - g(r)

    So, we have:

    Sum from n = 2 to N of Log(1-1/n^2) =

    Sum from n = 2 to N of
    Delta(Delta f)(n) =

    (Delta f)(N+1) - (Delta f)(2)

    Now

    f(n) = log(n-1),

    so

    (Delta f)(n) = Log(n) - log(n-1) =

    Log[n/(n-1)]

    and we see that:

    (Delta f)(N+1) - (Delta f)(2) =

    Log[(N+1)/N] - Log(2)


    The sum to infinity is the limit for N to infinity. We can write:

    Log[(N+1)/N] = Log(1+1/N) =
    1/N + O(1/N^2)

    So, this terms goes to zero in the limit N to infinity.

    The summation is thus equal to -Log(2)

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