Big O notation and Infinite series

Question

Suppose I know $\displaystyle\sum_{n\leq x}f(n)=O(g(x))$. Can I deduce that $\displaystyle\sum_{n\geq x}f(n)=O(g(x))$?

I think it should be true, and if anything, it should be a weak bound.

Answer 1

Such a bound does not hold. In fact, the tail of the (infinite) series can have a divergent sum.

Take $f(n) := n$. Hence, $$ \sum_{n=1}^x f(n) = \frac{x(x+1)}{2} = O(g(x)) $$ where $$ g(x) := x^2. $$ But $$ \sum_{n=x}^{+\infty} n = +\infty \neq O(g(x)). $$