True. For every integer , and , so
Thus and witness the required Big O notation bound.
False. Set . The ratio of successive terms is
Consequently for , which is unbounded. No constant can bound by that constant times for all sufficiently large . The factorial therefore fails the proposed Big O notation bound, without needing Stirling's formula.
True. Every term has when . Adding these inequalities gives
The constants and suffice in the definition of Big O notation.
False. The sum of squares of the first n positive integers gives
This ratio is unbounded, so it cannot satisfy the constant bound required by Big O notation. In fact the sum has order , as its leading term also shows.

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