To find the sum of the series \( \frac{2}{216} + \frac{2}{215} + \cdots + \frac{2}{2} + 2016 \), we can break it down into two parts:

1. The sum of the fractions: \( \frac{2}{216} + \frac{2}{215} + \cdots + \frac{2}{2} \)
2. The constant term: \( 2016 \)

### Sum of the Fractions

First, factor out the common factor of 2 from each term in the series:

\[ \frac{2}{216} + \frac{2}{215} + \cdots + \frac{2}{2} = 2 \left( \frac{1}{216} + \frac{1}{215} + \cdots + \frac{1}{2} \right) \]

The series inside the parentheses is a harmonic series starting from \( \frac{1}{2} \) to \( \frac{1}{216} \). The harmonic series \( H_n \) is defined as:

\[ H_n = \sum_{k=1}^{n} \frac{1}{k} \]

However, in this case, we need the sum from \( \frac{1}{2} \) to \( \frac{1}{216} \). This can be written as:

\[ \sum_{k=2}^{216} \frac{1}{k} \]

This is the harmonic number \( H_{216} \) minus the first term \( \frac{1}{1} \):

\[ \sum_{k=2}^{216} \frac{1}{k} = H_{216} - 1 \]

The harmonic number \( H_{216} \) can be approximated using the natural logarithm:

\[ H_{216} \approx \ln(216) + \gamma \]

where \( \gamma \) (the Euler-Mascheroni constant) is approximately 0.577.

So,

\[ H_{216} \approx \ln(216) + 0.577 \]

Using \( \ln(216) \approx 5.375 \):

\[ H_{216} \approx 5.375 + 0.577 = 5.952 \]

Thus,

\[ \sum_{k=2}^{216} \frac{1}{k} \approx 5.952 - 1 = 4.952 \]

Therefore,

\[ 2 \left( \sum_{k=2}^{216} \frac{1}{k} \right) \approx 2 \times 4.952 = 9.904 \]

### Adding the Constant Term

Now, add the constant term 2016:

\[ 9.904 + 2016 = 2025.904 \]

So, the sum of the series \( \frac{2}{216} + \frac{2}{215} + \cdots + \frac{2}{2} + 2016 \) is approximately \( 2025.904 \).