# Prove that $\sum\limits_{n=1}^{N} \frac{1}{n} \le \text{exp}\left(2\sum\limits_{n=1}^{N} p_k^{-1}\right)$

I want to proof that

$$\sum_{n=1}^{N} \frac{1}{n} \le \text{exp}(2\sum_{n=1}^{N} p_k^{-1})$$

where the sequence $$(p_n)_n$$ denotes the prime sequence.

I tried to do this by induction over N, but I am stuck at the induction step:

$$\sum_{n=1}^{N+1} \frac{1}{n} = \sum_{n=1}^{N} \frac{1}{n} + \frac{1}{N+1}$$

and by induction hypothesis we get:

$$\le \text{exp}(2\sum_{n=1}^{N} p_k^{-1}) + \frac{1}{N+1}$$

So now the aim should be to replace the term $$+ \frac{1}{N+1}$$ by $$\cdot \ \ \text{exp}(2 p_{N+1}^{-1})$$

This led me to the assumption that for any $$a > 1$$ holds:

$$a + \frac{1}{N} \le a \cdot \text{exp}(2 \frac{1}{p_N})$$

By the defintion of the exp function this is equivalent to

$$a + \frac{1}{N} \le a + a \cdot \sum_{k=1}^{\infty}\frac{2^k}{p_{N}^k \cdot k!}$$

and this can be reduced to:

$$\frac{1}{N} \le a \cdot \sum_{k=1}^{\infty}\frac{2^k}{p_{N}^k \cdot k!}$$

Unfortunately I do not know how to proof this assumption or even whether it is right or wrong. Could you help me?

## 1 Answer

Hint. Note that $$p_N\geq N$$ and, by fundamental theorem of arithmetic, any positive integer $$n\leq N$$ has a prime factorization involving powers of primes $$p_i$$ with $$1\leq i\leq N$$. Hence $$\sum_{n=1}^{N} \frac{1}{n}\leq \prod_{n=1}^N\left(1+\frac{1}{p_n}+\frac{1}{p_n^2}+\frac{1}{p_n^3}+\dots\right)=\prod_{n=1}^N\frac{1}{1-1/p_n}.$$ It remains to show that for $$x\in (0,1/2]$$, $$\frac{1}{1-x}\leq e^{2x}.$$ For this purpose, consider the derivative of the function $$f(x)=e^{2x}(1-x)$$.