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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.

Math Attack
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Shean
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  • I can think of Examples where it is true & Examples where it may not be true. It would be good to include some Example $f(n)$ & Example $g(x)$ in your Post , which will make the Query unambiguous. – Prem Apr 24 '23 at 18:43

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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 $$ \sum_{n=x}^{+\infty} n = +\infty \neq O(g(x)). $$

cs89
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  • In my earlier Comment , I alluded to other Cases. [ Eg : $f(n)=0$ , $f(n)=1$ , $f(n)=n$ , $f(n)=n(-1)^n$ , $f(n)=2^{-n}$ , $f(n)=2^{+n}$ ] Various Cases arise , though OP Question is not focussed. – Prem Apr 25 '23 at 10:12