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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

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

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The previous title gave me high hopes... oh well. –  J. M. Feb 10 '12 at 0:00
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1 Answer

up vote 7 down vote accepted

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 : $$B_1=\gamma+\sum_p \log\left(1-\frac1p\right)+\frac1p$$

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