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$$ show \sum_{k=2^n}^{2*2^n-1} \frac{1}{k} >= \frac{1}{2} $$
basic induction won't work because there is no overlap in ks for n+1
$$ \sum_{k=2^{n+1}}^{2*2^{n+1}-1} \frac{1}{k} = \sum_{k=2*2^{n}}^{4*2^{n}-1} \frac{1}{k} $$

we can write dividend and divisor as separate sequences
but this doesn't seem productive
$$ \sum_{k=2^n}^{2*2^n-1} \frac{1}{k} = \frac{\sum_{j=2^n}^{2*2^j-1} \frac{\prod_{k=2^n}^{2*2^n-1} k}{j}} {\prod_{k=2^n}^{2*2^n-1} k} $$

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my best attempt:
$$ show \sum_{k=2^n}^{2*2^n-1} \frac{1}{k} >= \frac{1}{2} $$
split into odd and even. above iterates over 2^n elements. following split will iterate over 2^n/2 elements twice
$$ \sum_{k=2^n}^{2*2^n-1} \frac{1}{k} = \sum_{k=2^{n-1}}^{2^n-1} \frac{1}{2k+1} + \sum_{k=2^{n-1}}^{2^n-1} \frac{1}{2k} $$
therefore
$$ \sum_{k=2^{n+1}}^{2*2^{n+1}-1} \frac{1}{k} = \sum_{k=2^{n}}^{2*2^n-1} \frac{1}{2k+1} + \frac{1}{2} \sum_{k=2^{n}}^{2*2^n-1} \frac{1}{k} $$
right-hand side looks familiar. "+1" part of 2k+1 gets exceedingly irrelevant
$$ \lim_{n \to \infty} (\sum_{k=2^{n}}^{2*2^n-1} \frac{1}{2k+1} - \frac{1}{2} \sum_{k=2^{n}}^{2*2^n-1} \frac{1}{k}) = 0 $$
and the following becomes true
$$ \lim_{n \to \infty} (\sum_{k=2^{n+1}}^{2*2^{n+1}-1} \frac{1}{k} - (\sum_{k=2^{n}}^{2*2^n-1} \frac{1}{2k+1} + \frac{1}{2}\sum_{k=2^{n}}^{2*2^n-1} \frac{1}{k})) = 0$$
if we can substitute one for the other (2k+1 part with 2k part)
$$ \lim_{n \to \infty} (\sum_{k=2^{n+1}}^{2*2^{n+1}-1} \frac{1}{k} - 2 * \frac{1}{2}\sum_{k=2^{n}}^{2*2^n-1} \frac{1}{k}) = 0$$
showing convergence

n=0 into original inequation gives 1 >= 1/2
    n=1 gives 1/2+1/3 >= 1/2
    n=2 gives 1/4+1/5+1/6+1/7 = (1/2+1/3)/2 + 1/5+1/7 >= 1/2

we have shown convergence. still to show that its limit is >= 1/2

seems obvious that it is, but how to actually demonstrate it?

n=0  1                          =     1         red-factor
    n=1  1/2 *   1     +    2/    3 =     5/    6   0.83333333333333333333333333333333
    n=2  1/2 *   5/  6 +   12/   35 =   319/  420   0.91142857142857142857142857142857
    n=3  1/2 * 319/420 + 6672/19305 = 52279/72072   0.95503503810713842061804444249585

red-factor gives the factor by which the numerical solution for n-1 must be multiplied to get the solution of n. we are converging fairly quickly and speed is bound to increase as the difference between 1/(2k+1) and 1/2k decreases

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seems like a decent argument, but i'm not satisfied with this

Thu, 18 Oct 2018 14:34 GMT