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$\log2 = \sum_{n=1}^k\frac{(-1)^{n-1}}{n} + (-1)^k2\int_{0}^1\frac{t^{2k+1}}{1+t^2}dt$

I'm having trouble even getting the base case. Any help would be appreciated.

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    @Thomas: Good point, I removed my comment because using it would result in circularity and so on.2012-12-12

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HINT: The base case is trivial if you start at $k=0$. If you insist on starting at $k=1$, divide out the integrand: $\int_0^1\frac{t^3~dt}{1+t^2}=\int_0^1\left(t-\frac{t}{1+t^2}\right)dt\;;$ then the integration and evaluation are easy. And this suggests how to handle the induction step:

$\frac{t^{2k+3}}{1+t^2}=\frac{(1+t^2)t^{2k+1}-t^{2k+1}}{1+t^2}=t^{2k+1}-\frac{t^{2k+1}}{1+t^2}\;.$

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    @Hal Yes, an empty sum is alwasys zero.2012-12-13
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For the base case, the sum is 1. To evaluate the integral, notice that $\frac{t^3}{t^2+1}=t-\frac{t}{t^2+1}$. Then the integral evaluates very easily to $(1-\log2)$, so we have $\log2=1-(1-\log2)=\log2$.

For the inductive step, simply show that $\frac{1}{k}=2\int_{0}^1\frac{t^{2k+3}-t^{2k+1}}{1+t^2}dt$. If I am not mistaken, you can evaluate the integral fairly easily.

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    Ah, yes, you are right.2012-12-13