Task
The number \pi can be defined through various infinite sums. The accuracy increases with the number of terms. Considering the following sum, which we call sum 2:
\frac{\pi}{2} = 1 + \sum_{j = 1}^{m - 1} \frac{\prod_{i=1}^j i}{\prod_{i=1}^j (2i + 1)} = 1 + \frac{1}{3} + \frac{1 \cdot 2}{3 \cdot 5} + \frac{1 \cdot 2 \cdot 3}{3 \cdot 5 \cdot 7} + ...
Write a program that computes and outputs an approximation of Pi, based on sum 2. The input for your program is the number of terms m (including 1) to be included in the calculation to be done. The output is the approximation of \pi.
Optional: After you have solved this and the previous task, think about which formula gives a more precise approximation of , sum 1 or sum 2 ? What are the drawbacks? Write your thoughts in a comment below the code.
Input
A natural number m (m \geq 1).
Output
The approximation of \pi given by m terms, rounded to 6 significant digits. Note that 6 significant digits is the default precision of C++ for printing floating-point values. Use a double variable to store the sum.
Important: the use of functions from the math library (e.g., pow) is prohibited.