# Copyright (C) 2020 Andreas Kloeckner # #

# ## MIT License

# Permission is hereby granted, free of charge, to any person obtaining a copy
# of this software and associated documentation files (the "Software"), to deal
# in the Software without restriction, including without limitation the rights
# to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
# copies of the Software, and to permit persons to whom the Software is
# furnished to do so, subject to the following conditions:
#
# The above copyright notice and this permission notice shall be included in
# all copies or substantial portions of the Software.
#
# THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
# IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
# FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
# AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
# LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
# OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
# THE SOFTWARE.
#

# You may know from math that
# $$
# \sum_{n=1}^\infty \frac 1n=\infty.
# $$
# Let's see what we get using floating point:
# In[1]:
import numpy as np
# In[2]:
n = int(0)
float_type = np.float32
my_sum = float_type(0)
while True:
n += 1
last_sum = my_sum
my_sum += float_type(1 / n)
if n % 200000 == 0:
print("1/n = %g, sum0 = %g"%(1.0/n, my_sum))
# In[ ]:
# In[ ]: