A limit that involves prime numbers 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 $\lim_{n\to\infty} a_{n} = \infty$. 
 A: 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. 
A: 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.
A: 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.
