I am trying to figure out how to prove $- \frac{\zeta'}{\zeta}(s) = F(s) -G(s)\zeta'(s) -F(s)G(s)\zeta(s) -(\zeta(s)G(s) -1) \Big{(} - \frac{\zeta'}{\zeta}(s) - F(s) \Big{)}$ to prove $\Lambda(n) = a_1(n) + a_2(n) + a_3(n) + a_4(n)$ as defined in this article. https://en.wikipedia.org/wiki/Vaughan's_identity
I know $- \frac{\zeta'}{\zeta}(s) = \sum_{n = 1}^\infty \frac{\Lambda(n)}{n^s}$, and $-\zeta'(s) = \sum_{n = 1}^\infty \frac{log(n)}{n^s}$, and $\zeta(s) = \sum_{n = 1}^\infty \frac{1}{n^s}$. So by substitution we have
$\sum_{n = 1}^\infty \frac{\Lambda(n)}{n^s}= F(s) -G(s)\zeta'(s) -F(s)G(s)\zeta(s) -(\zeta(s)G(s) -1) \Big{(} - \frac{\zeta'}{\zeta}(s) - F(s) \Big{)}$ $= F(s) +G(s)\sum_{n = 1}^\infty \frac{log(n)}{n^s} -F(s)G(s)\sum_{n = 1}^\infty \frac{1}{n^s} -(G(s)\sum_{n = 1}^\infty \frac{1}{n^s} -1) \Big{(} \sum_{n = 1}^\infty \frac{\Lambda(n)}{n^s} - F(s) \Big{)}$ $= \sum_{m \leq u} \frac{ \Lambda(m)}{m^s} +G(s)\sum_{n = 1}^\infty \frac{log(n)}{n^s} -(\sum_{m \leq u} \frac{ \Lambda(m)}{m^s})G(s)\sum_{n = 1}^\infty \frac{1}{n^s} -(G(s)\sum_{n = 1}^\infty \frac{1}{n^s} -1) \Big{(} \sum_{n = 1}^\infty \frac{\Lambda(n)}{n^s} - \sum_{m \leq u} \frac{ \Lambda(m)}{m^s} \Big{)}$ $= \sum_{m \leq u} \frac{ \Lambda(m)}{m^s} +(\sum_{m \leq v} \frac{ \mu(m)}{m^s})\sum_{n = 1}^\infty \frac{log(n)}{n^s} -(\sum_{m \leq u} \frac{ \Lambda(m)}{m^s})(\sum_{m \leq v} \frac{ \mu(m)}{m^s})\sum_{n = 1}^\infty \frac{1}{n^s} -(\sum_{m \leq v} \frac{ \mu(m)}{m^s}\sum_{n = 1}^\infty \frac{1}{n^s} -1) \Big{(} \sum_{n = 1}^\infty \frac{\Lambda(n)}{n^s} - \sum_{m \leq u} \frac{ \Lambda(m)}{m^s} \Big{)}$
**edit to add
$= \sum_{n \leq u} \frac{ \Lambda(n)}{n^s} +(\sum_{d \leq v} \frac{ \mu(d)}{d^s})\sum_{h = 1}^\infty \frac{log(h)}{h^s} -(\sum_{m \leq u} \frac{ \Lambda(m)}{m^s})(\sum_{d \leq v} \frac{ \mu(d)}{d^s})\sum_{n = 1}^\infty \frac{1}{n^s} -(\sum_{d \leq v} \frac{ \mu(d)}{d^s}\sum_{n = 1}^\infty \frac{1}{n^s} -1) \Big{(} \sum_{m = 1}^\infty \frac{\Lambda(m)}{m^s} - \sum_{m \leq u} \frac{ \Lambda(m)}{m^s} \Big{)}$.
$= \sum_{n \leq u} \frac{ \Lambda(n)}{n^s} +(\sum_{n = 1}^\infty (\sum_{dh = n, d \leq v}\mu(d)log(h))n^{-s} -\sum_{n = 1}^\infty(\sum_{md |n, m \leq u, d \leq v} \Lambda(m)\mu(d)) \frac{1}{n^s} -(\sum_{n = 1}^\infty(\sum_{d | n, d \leq v} \mu(d)) n^{-s} -1) \Big{(} \sum_{m > u}^\infty \frac{\Lambda(m)}{m^s} \Big{)}$
$= \sum_{n \leq u} \frac{ \Lambda(n)}{n^s} +(\sum_{n = 1}^\infty (\sum_{dh = n, d \leq v}\mu(d)log(h))n^{-s} -\sum_{n = 1}^\infty(\sum_{md |n, m \leq u, d \leq v} \Lambda(m)\mu(d)) \frac{1}{n^s} -\sum_{n = 1}^\infty(\sum_{mk = n, m > u, k > v} \Lambda(m)\sum_{d | k, d \leq v} \mu(d)) n^{-s} + \sum_{m > u}^\infty \frac{\Lambda(m)}{m^s} $
The only problem is in order to get what I want by comparing the coefficients of $n^{-s}$ I need to show $\sum_{m > u}^\infty \frac{\Lambda(m)}{m^s} = 0$