Calculate pi using the Nilakantha series with up to 48 decimals of precision

pi.py

'''
calculate pi using the Nilakantha series up to 48 decimals of precision
Nilakantha series approximates pi, but is not as accurate as some other, more intensive methods
the most accurate predictions come with odd number decimal precision
''

def calculate_pi(decimals):

	# make the calculation
	
	pi = 0
	for i in range(1,decimals+1):

		# All this weirdness is part of the Nilikantha series

		change = 4 / ((i * 2) * ((i * 2) + 1) * ((i * 2) + 2))
		
		if (i % 2 == 0):
			pi -= change
		else:
			pi += change
			
	pi = pi + 3

	# cut answer off at required decimals

	pi = f"%.{decimals}f" % pi
	
	# print results

	return f"\nPi calculated to {decimals} decimals of precisions is:\n{pi}\n\n\n"

# get required decimals of precision

while True:
	
	# ask for decimals of precision (Nilakantha series is most accurate with odd numbers)

	decimals = input("\nHow many decimals of precision do you need? > ")
	
	# validate that it's a number

	try:
		int(decimals)
	except:
		print("Error: Please enter a whole number.")
		continue
		
	decimals = int(decimals)

	# max out at 1000 decimal places
		
	if (decimals > 48):
		print(f"Error: {decimals} is too large. I automatically set it to 48 for you.")
		decimals = 48
		
	# require at least 2 decimal places
		
	if (decimals < 2):
		print(f"Error: {decimals} is too small. I automatically set it to 2 for you.")
		decimals = 2
				
	# fire up the calculator
	
	print(calculate_pi(decimals))
	
	# end the while loop
	
	break