Does $g(n)=O(f(n))$ imply $\sum_n {g(n)} = O(\sum_n {f(n))}$?

Question

Let be $g(n)=O(f(n))$ it is true that $$\sum\limits_n {g(n)} = O(\sum\limits_n {f(n))}$$ when f and g are arithmetic functions. and specifically $$\sum\limits_{d|p - 1} {\mu (d)} O(x{p^{{\textstyle{1 \over 2}}}}) =O(\sum\limits_{d|p - 1} {\mu (d)x{p^{{\textstyle{1 \over 2}}}})}= O({2^m}x{p^{{\textstyle{1 \over 2}}}})?$$ when $x$ is real, $p$ is a prime number, $\mu$ is mobius function and $m$ is the number of distinct primes that divide $p-1$?

Answer 1

For $n\ge N$, $g(n)<c f(n)$ allows you to write that for $n\ge N$, $\displaystyle\sum_{k=N}^n g(k)<c\sum_{k=N}^n f(k)$, which indeed expresses that

$$\sum_{k=N}^n g(k)=O\left(\sum_{k=N}^n f(k)\right).$$

---

If the lower bound of the summation precedes $N$, let $0$, we can adapt with

$$\sum_{k=0}^n g(k)<\sum_{k=0}^{n-1} g(k)+c\sum_{k=N}^n f(k)=\sum_{k=0}^{n-1} g(k)+c\sum_{k=0}^{n-1} f(k)+c\sum_{k=N}^n f(k)$$

and

$$\sum_{k=0}^n g(k)=O\left(\sum_{k=N}^n f(k)+A\right)=O\left(\sum_{k=0}^n f(k)+B\right)$$ where $A,B$ are constants.

This only makes a difference in case of an $o(1)$ behavior.