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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?

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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 $f(x)=e^{2x}(1-x)$.

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