How prove $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\cdots+\frac{1}{a_{n}}<2$

Let $$A=\{a_{1},a_{2},\ldots,a_{n}\}\subset N$$ Suppose that for any two distinct subsets $B, C\subseteq A$, we have $$\sum_{x\in B}x\neq \sum_{x\in C}x$$

Then show that $$\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\dfrac{1}{a_{3}}+\cdots+\dfrac{1}{a_{n}}<2$$

This is an old conjecture of Erdos, which was subsequently proved (Ryavek from my notes) - though I cannot find a handy online reference just now. The proof goes along the following lines, IIRC:

With $0 < x< 1$, we have by the distinct sum condition: $$ \prod_{k=1}^n (1+x^{a_k}) < \sum_{k=0}^{\infty} x^k = \frac1{1-x} $$

$$\implies \sum_{k=1}^n \log(1+x^{a_k}) < - \log (1-x)$$

As both sides are positive, we can divide by $x$ and integrate to get

$$\implies \sum_{k=1}^n \int_0^1 \frac{ \log(1+x^{a_k})}x dx < - \int_0^1 \frac{\log (1-x)}x dx $$

$$\implies \sum_{k=1}^n \frac1{a_k} \cdot \int_0^1 \frac{ \log(1+t)}t dt < \frac{\pi^2}6 $$ $$\implies \sum_{k=1}^n \frac1{a_k} \cdot \frac{\pi^2}{12} < \frac{\pi^2}6 \implies \sum_{k=1}^n \frac1{a_k} < 2 $$