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I know that $\sum_{p \leq N} \frac{1}{p} \geq \log\log N -1 $

However, want to show that $\sum_{p \leq x} \frac{1}{p} \geq \log\log x -1 $.

If let $N=[x]$, then we get a bound for x, i.e. $N \leq x <N+1$. However, from that all I can seem to get is this $\sum_{p \leq N+1} \frac{1}{p} \geq \log\log(N+1) -1 >\log\log x-1 $

Can't seem to be able to conclude that. As we have $\frac{1}{N+1}$

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You are right, the assertion with $x$ is a very tiny bit stronger than the assertion with $N$. It would certainly be possible to find a function $f$ such that $f(N) \ge \log\log N-1$ but $f(N) <\log\log (N+1/2) -1$. One would have to look back on the proof of the result for $N$ and see what kind of slack there is. Since $\log\log$ grows with icy slowness, not much slack is needed. –  André Nicolas Feb 9 '12 at 1:24
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Your post seems to have died in mid-air. Anyway, maybe you want to look beyond the fact $\sum^N\ge\log\log N-1$ and study a proof of the fact and see whether the proof can be made to work for $\sum^x$. –  Gerry Myerson Feb 9 '12 at 1:25
    
I cleaned up the TeX code. Please don't write $log log N$ if you mean $\log\log N$. The code for the former is "log log N"; for the latter it's "\log\log N". –  Michael Hardy Feb 9 '12 at 3:49

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Well, in your other thread we got $\displaystyle \sum_{p \leq N} \frac{1}{p} \geq \log\log(N+1) -1\ \ $ (with $N \to N+1$)

and this is enough to get $\displaystyle \sum_{p \leq x} \frac{1}{p} \geq \log\log x -1\ \ $ (for $N=\lfloor x \rfloor$)

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Thanks for that. I was doing something really stupid. You have been really helpful again. –  simplicity Feb 9 '12 at 1:35
    
@simplicity: I have a doubt now, it seems we got rather $\displaystyle \sum_{p \leq N} \frac{1}{p-1} \geq \log\log(N+1)\ $ instead... I'll have to look at this again tomorrow... –  Raymond Manzoni Feb 9 '12 at 1:53
    
@simplicity: I updated my answer in the other thread to get your initial inequality. –  Raymond Manzoni Feb 9 '12 at 23:51

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