Evaluate the sum

$$
\sum_{k=1}^{\infty}\left(\prod_{i=1}^{k} \frac{P_{i}-1}{P_{i+1}}\right)=\frac{1}{3}+\frac{1}{3} \cdot \frac{2}{5}+\frac{1}{3} \cdot \frac{2}{5} \cdot \frac{4}{7}+\frac{1}{3} \cdot \frac{2}{5} \cdot \frac{4}{7} \cdot \frac{6}{11}+\ldots
$$

where $P_{n}$ denotes the $n^{\text {th }}$ prime number.
Rewrite the given sum as

$$
\begin{gathered}
\frac{1}{3}+\frac{2}{3} \cdot \frac{1}{5}+\frac{2}{3} \cdot \frac{4}{5} \cdot \frac{1}{7}+\frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7}+\cdot \frac{1}{11}+\ldots \\
=\frac{1}{3}+\left(1-\frac{1}{3}\right) \frac{1}{5}+\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right) \frac{1}{7}+\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right) \frac{1}{11}+\ldots \\
=\frac{1}{3}+\frac{1}{5}-\frac{1}{3 \cdot 5}+\frac{1}{7}-\frac{1}{3 \cdot 7}-\frac{1}{5 \cdot 7}+\frac{1}{3 \cdot 5 \cdot 7}+\frac{1}{11}-\frac{1}{3 \cdot 11} \frac{1}{5 \cdot 11}-\frac{1}{7 \cdot 11}+\ldots
\end{gathered}
$$

Note that each term in this sum must have denominator of the form $p_{1} p_{2} \ldots p_{k}$, where the $p_{i}$ are distinct primes (WLOG $p_{1}<p_{2}<\cdots<p_{k}$ ). Further, each denominator
occurs exactly once, namely, in expanding the product whose last term is $\frac{1}{p_{k}}$, and its sign is $(-1)^{k-1}$, since it comes from the term

$$
\left(-\frac{1}{p_{1}}\right)\left(-\frac{1}{p_{2}}\right) \ldots\left(-\frac{1}{p_{k-1}}\right) \frac{1}{p_{k}}
$$

in the expansion of the product containing $\frac{1}{p_{k}}$. However, this lets us rewrite the left side as

$$
1-\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right) \ldots
$$

However, we can see that

$$
\begin{gathered}
\frac{1}{\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)}=\frac{1}{1-\frac{1}{3}} \cdot \frac{1}{1-\frac{1}{5}} \cdot \frac{1}{1-\frac{1}{7}} \ldots \\
=\left(1+\frac{1}{3}+\frac{1}{3^{2}}+\ldots\right)\left(1+\frac{1}{5}+\frac{1}{5^{2}}+\ldots\right)\left(1+\frac{1}{7}+\frac{1}{7^{2}}+\ldots\right)=\sum_{n=1}^{\infty} \frac{1}{2 n+1} .
\end{gathered}
$$

Thus, since the Harmonic series diverges, the product

$$
\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right) \ldots
$$

must equal 0 , so the desired sum is $1-0=1$.