How to tell the system that the service has started?

Question

linux

Hello, I want to run a python script as a service, the script starts and runs, but the system shows it in the activating state, which is why it constantly restarts, maybe

systemd[1]: rpi-rf.service: Failed with result 'timeout'.

.
How to tell the system that the script has started and is running?
.service:

# systemd unit file for the Python Demo Service


[Unit]

# Human readable name of the unit
Description=Python rpi-rf_receiver


[Service]

# Command to execute when the service is started
ExecStart=/usr/bin/python3 /usr/local/lib/rpi-rf/rpi-rf_receive

# Disable Python's buffering of STDOUT and STDERR, so that output from the
# service shows up immediately in systemd's logs
Environment=PYTHONUNBUFFERED=1

# Automatically restart the service if it crashes
Restart=on-failure
RestartSec=5

# Our service will notify systemd once it is up and running
Type=notify

# Use a dedicated user to run our service
User=root


[Install]

# Tell systemd to automatically start this service when the system boots
# (assuming the service is enabled)
WantedBy=default.target

Answer 1

A more correct way

from systemd.daemon import notify, Notification

# Send READY=1
notify(Notification.READY)

Answer 2

Change Type=notifytoType=simple

Answer 3

Why is the Type=notify option added?