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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 $p-1$?

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    Do you mean indefinite summation by $\sum_n$? Cause if not, that won't work as you intend it to, because it would just mean something like $5=O(21)$, because you evaluate your functions. The usual definitions of $O$ with $n$ being taken as indeterminate wouldn't work because it is evaluated. – 2017-01-04
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    Also it is unclear what $\mu(d)$ and $m$ are in your second equation – 2017-01-04
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    @Ayutac please explain both cases, finite and infinite – 2017-01-04

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For $n\ge N$, $g(n)

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

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.

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    how about second part? – 2017-01-04
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    @alexmorfi: use my method. – 2017-01-04
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    please look at this question maybe you can help me http://math.stackexchange.com/questions/2076742/bruns-method-and-primitive-roots – 2017-01-04
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    @alexmorfi: sorry, I don't work "on demand". – 2017-01-04