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Let be $p_{n}$ the nth prime number and $(a_{n}), n\geq1$ such that: $$a_{n}=\frac{1}{p_1}+\frac{1}{p_2}+\cdots+\frac{1}{p_n}$$

By using this result, $$\lim_{n\rightarrow\infty} \frac{p_{1}}{p_{1}-1} \frac{p_{2}}{p_{2}-1}\cdots\frac{p_{n}}{p_{n}-1}=\infty$$ I have to prove that $\lim_{n\to\infty} a_{n} = \infty$.

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Wait. I saw something before and wrote the answer accordingly. – user17762 Jun 11 '12 at 17:18
Thanks for all your answers. – user 1618033 Jun 11 '12 at 19:42
up vote 2 down vote accepted

Absolute convergence of $\displaystyle \prod_{k=1}^{\infty} \left( 1+a_k\right)$ means that $\displaystyle \sum_{k=1}^{\infty} \lvert a_k \rvert$ converges.

Further if $a_k$'s are positive, then we have the following inequalities. $$1 + \sum_{k=1}^{\infty} a_k \leq \displaystyle \prod_{k=1}^{\infty} \left( 1+a_k\right) \leq \exp \left( \sum_{k=1}^{\infty} a_k \right)$$ Hence, if $a_k$'s are positive, then $\displaystyle \prod_{k=1}^{\infty} \left( 1+a_k\right)$ converges iff $\displaystyle \sum_{k=1}^{\infty} a_k$ converges.

Since we are given that $\displaystyle \prod_{k=1}^{\infty} \left(\dfrac{p_k}{p_k-1} \right) = \prod_{k=1}^{\infty} \left(1 + \dfrac1{p_k-1} \right)$ diverges, we have that $$\sum_{k=1}^{\infty} \dfrac1{p_k-1}$$ diverges i.e. if $b_n = \displaystyle \sum_{k=1}^{n} \dfrac1{p_k-1}$, then $\displaystyle \lim_{n \rightarrow \infty} b_n = \infty$.

Now note that \begin{align} a_n & = \dfrac1{p_1} + \dfrac1{p_2} + \dfrac1{p_3} + \cdots + \dfrac1{p_n} > \dfrac1{p_2-1} + \dfrac1{p_3-1} + \dfrac1{p_4-1} + \cdots + \dfrac1{p_n-1} + \dfrac1{p_{n+1}-1}\\ & = b_{n+1} - \dfrac1{p_1-1} = b_{n+1} - 1 \end{align} The above is true since successive prime differ at-least by $1$ i.e. $p_k < p_{k+1}-1$.

Now you letting $n \to \infty$, we get what you want.

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I think you meant in your first definition absolute convergence, @Marvis. Otherwise one has to drop the absolute value from the series. – DonAntonio Jun 11 '12 at 18:45
@DonAntonio Thanks for pointing it out. I have now edited it accordingly. – user17762 Jun 11 '12 at 18:51

Hint: Each term of your product can be expressed in the shape $\dfrac{1}{1-\frac{1}{p}}$.

Take the logarithm of your product, and use an estimate for $\ln(1+x)$ when $|x|$ is close to $0$.

Added: To complete the proof, note that $\ln\left(\frac{1}{1-x}\right)=-\ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots$ when $|x|$ is small. It follows that in particular when $x$ is small positive, $\ln\left(\frac{1}{1-x}\right)<2x$. Put $x=\frac{1}{p_k}$. We find by Comparison that $\sum_1^\infty \frac{1}{p_k}$ diverges.

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thanks for your good idea. I think of some direct result that mainly involves that product. – user 1618033 Jun 11 '12 at 17:29

This isn't a direct answer to your question, but it's a nice way to convince yourself that $a_n$ diverges, so I wanted to include it. The prime number theorem implies that the $n$th prime is roughly $n\log n$, so your sum can be approximated by:

$$a_n \approx\int_e^n \frac{1}{x\log x}dx = \int_e^n \frac{(1/x)}{\log x}dx = \left[\log\log x\right]_e^n = \log\log n$$

which clearly diverges.

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yeah. This is very helpful and clear when one wants to prove the divergence. – user 1618033 Jun 11 '12 at 19:43

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