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I can't see how you get this.

I want to show that

$\log(\log(N+1)) \leq \sum_{p \leq N} \frac{1}{p}+1$

Can't see how it follows from this. So you show that $0 \lt -\log(1-x)-x \lt \frac{x^2}{(1-x)}$.

Which is fine, but then the lecturer does a jump I don't understand.

He rewrites this $\sum_{p \leq N} \frac{1}{p} + \sum_{p\leq N} (-\log(1-\frac{1}{p}) - \frac{1}{p})$

Know that
$ \sum_{p\leq N} (-\log(1-\frac{1}{p}) - \frac{1}{p}) \leq \sum_{p \leq N} \frac{1}{p(p-1)}$

Then just claims that this shows

$\log(\log(N+1)) \leq \sum_{p \leq N} \frac{1}{p} +1$

  • 1
    The previous title gave me high hopes... oh well.2012-02-10

1 Answers 1

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Let's start with : $ \sum_{p\leq N} \left(-\log\left(1-\frac{1}{p}\right) - \frac{1}{p}\right) \leq \sum_{p \leq N} \frac{1}{p(p-1)}$

The idea is to use the finite version of Euler's product : $\prod_{p\le N} \frac1{1-p^{-1}} =\prod_{p\le N} \left(1+\frac{1}{p}+\frac{1}{p^2}+\cdots\right)=\sum_{n\in S_N} \frac1n$
with $S_N$ the set of integers composed of prime factors $p\le N$

Since this set includes all the integers $n\le N$ we have :
$\prod_{p\le N} \frac1{1-p^{-1}} \ge H_N$

but the harmonic sum $H_N$ is well known to be greater than $\log(N+1)$ (use for example minoration by $\int_1^{N+1} \frac1n dn$).

So that (taking logarithms) we get : $\log(\log(N+1)) \le \sum_{p\le N} -\log\left(1-\frac1p\right)$ that we may replace in your inequality.


UPDATE2: Let's handle with more care this replacement to show that indeed : $\log(\log(N+1)) \leq \sum_{p \leq N} \frac{1}{p} +1$

from the convergent : $\ \displaystyle 0 \le \sum_{p\leq N} \left(-\log\left(1-\frac{1}{p}\right) - \frac{1}{p}\right) \leq \sum_{p \leq N} \frac{1}{p(p-1)}$
we get (adding the sum on $\frac1p$) : $ \sum_{p\leq N} \frac{1}{p}\le \sum_{p\leq N} -\log\left(1-\frac{1}{p}\right) \leq \sum_{p \leq N} \frac{1}{p-1}$ At this point we may use the previous result to get not your inequality but : $ \log(\log(N+1)) \le \sum_{p \leq N} \frac{1}{p-1}$

Let's use the fact that $\frac1{p-1}\le \frac1q$ (if $q$ is the prime before $p$) or $1$ (for $p=2$) to rewrite this as ('shifting' the primes and extracting $1$) : $ \log(\log(N+1)) \le 1+\left(\sum_{p \le N} \frac1p\right)-\frac1{p_N}\ ,\ \text{with }p_N\ \text{the largest prime}\ \le N$ this implies indeed : $ \log(\log(N+1)) \le 1+\sum_{p \le N} \frac1p$

We may replace the $1+$ term by $\frac12+$ by noticing that your initial inequality could have been (adding $2$ at the denominator) : $0 \lt -\log(1-x)-x \lt \frac{x^2}{2(1-x)}$ because the effect of this is to replace the majoration of $\frac1{p(p-1)}+\frac1p=\frac1{p-1}$ by $\frac1{2p(p-1)}+\frac1p=\frac{p+p-1}{2p(p-1)}=\frac12(\frac1{p-1}+\frac1p)$ so that the final inequality is replaced by : $ \log(\log(N+1)) \le \frac12\left(1+\sum_{p \le N} \frac1p+\sum_{p \le N} \frac1p\right)$


We could continue this way but better results may be proved (see for example the excellent Hardy&Wright 'An introduction to the theory of numbers')
using the function $\displaystyle C(x)=\sum_{p\le x} \frac{\log p}p$ and the equality $\displaystyle \sum_{p\le x}\frac1p=\frac{C(x)}{\log x}+\int_2^x \frac{C(t)}{t\ \log^2(t)}dt\ $ to get : $\sum_{p\le x}\frac1p= \log\log x+B_1+o(1)$ with $B_1$ the Merten's constant : $B_1=\gamma+\sum_p \log\left(1-\frac1p\right)+\frac1p$