I would like to ask about a simple example of calculation of a zero for some partial sum $\sum_{n=1}^N\frac{\mu(n)}{n^s}$ of the Dirichlet series $$\sum_{n=1}^\infty\frac{\mu(n)}{n^s},$$ where $\mu(n)$ is the Möbius function. I am interested in learn with an example for the case $\Re s>1$.
Our example will be taking $N=7$, then we define $$f(s)=\sum_{n=1}^7\frac{\mu(n)}{n^s}=1-2^{-s}-3^{-s}-5^{-s}+6^{-s}-7^{-s}.$$ On assumption that there exists a zero $\rho$, $f(\rho)=0$ with $\Re s>1$ (I did calculations with Wolfram Alpha with code sum mu(n)/n^s, from n=1 to 7 you can get our expression, and after with zeros 1-2^(-s) - 3^(-s) - 5^(-s) + 6^(-s) - 7^(-s) shows a possible zero) then taking the real, and imaginary parts implementing our condition to be a zero $\rho=x+iy$
$$2^{-x}\cos(y\log 2)+3^{-x}\cos(y\log 3)+5^{-x}\cos(y\log 5)-6^{-x}\cos(y\log 6)+7^{-x}\cos(y\log 7)=1$$ and $$2^{-x}\sin(y\log 2)+3^{-x}\sin(y\log 3)+5^{-x}\sin(y\log 5)-6^{-x}\sin(y\log 6)+7^{-x}\sin(y\log 7)=0.$$
Question. I don't know if by means of my calculations you can dilucidate something. What is your approach to show that our $$f(s)=\sum_{n=1}^7\frac{\mu(n)}{n^s}$$ has a zero with $\Re s>1$? Only is required a proof for the existence of such zero. Many thanks.
If you answer the question using a numerical method implemented in your computer, please illustrate how works your method mathematically, that is in whatare the rules of numerical analysis that you are using.
I don't know if taking different examples of $N$ one can define different functions having zeros with $\Re \rho>1$ and also $\Im \rho\neq 0$.
Hint: Notice that $f(1)=-\frac1{105}$ and $f(2)=\frac{2228}{3675}$. Use the Intermediate Value Theorem.