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After I've read a question and I've understand its corresponding answer of a post recently edited in this site Mathematics Stack Exchange I tried do calculations concerning the definition of $$f(s)=\sum_{n=1}^\infty\frac{\mu(n)-\mu(n+1)}{n^s}.$$ See the definition and where is convergent in previous reference. My approach to do my calculations (are a set of simple manipulations to get an artificious identity) was to combine an identity involving the Mertens function $M(n)$ and the Möbius function $\mu(n)$, see the remarks below next question.

Question. A) If we assume that our series converge absolutely, say us for this calculations for $\Re s>\sigma_a$, is it right that $$1-f(s)=S(s)+T(s),$$ where $$S(s)= \sum_{n=1}^{\infty} \left[ -\frac{M(n)}{(n+1)n^{s+1}} + \left( \frac{1}{(n+1)^s}-\frac{1}{n^s}\right)\sum_{k=1} ^{n-1}\frac{M(k)}{k(k+1)}\right] $$ and $$T(s)= \sum_{n=1}^{\infty} \left[ \left( \frac{n+1}{n^s}-\frac{1}{(n+1)^{s-1}} \right)\frac{\mu(n+1)}{n+1} + \left( \frac{1}{n^s}+\frac{1}{(n+1)^s}\right)\sum_{k=1} ^{n}\frac{\mu(k)}{k}\right] ? $$

B) What's easy $\sigma_a$ in the condition for the convergence $\Re s>\sigma_a$ can be stated? Thanks in advance.

Thus I am asking A) as a proof verification, for which I am waiting if there are mistakes that you tell me or provide me your final statement. As I said I've deduced by means of simple manipulations when I've written $$\sum_{n=1}^\infty(M(n+1)-M(n))\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right)$$ and try to combine with the definiton of $f(s)$ and the identity that satisfy the Mertens and Möbius functions, that is (1.8) of page 2 from [1] that can find for example in (currently available in arXiv with article ID arXiv:math/0011254). For B) I am asking a good (only reasonings to get an approximation) abscissa showing the absolute convergence for all manipulations that you need (I know unconditional asymptotics for the functions that involve previous equations, thus is only required hints or a guideline to get your claim).

References:

[1] Báez-Duarte, Arithmetical Aspects of Beurling's Real Variable Reformulation of the Riemann Hypothesis, (2000).

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    I tried to assess your equation numerically, but couldn't reconcile the numbers. Are you sure the formula is typed correctly? Using your arxiv reference, I could derive the following: $$f(1)=\sum_{k=2}^{\infty} \left(1-\frac{1}{\zeta(k)}\right)=\sum_{n=1}^{\infty} \frac{M(n)-M(n+1)}{n(n+1)}$$ Have not yet found a way to link the RHS back to an $f(s)$ type function. – 2017-01-16
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    These are my new calculations, but I don't know $100$% if were rights$$\hat{S}(s)= \sum_{n=1}^{\infty} \left[ \left(\frac{1}{(n+1)^{s-1} }-\frac{n+1}{n^s} \right) \frac{M(n)}{n(n+1)} + \left( \frac{1}{(n+1)^s}-\frac{1}{n^s}\right)\sum_{k=1} ^{n-1}\frac{M(k)}{k(k+1)}\right],$$ and $$\hat{T}(s)= \sum_{n=1}^{\infty} \left[ \left( \frac{n+1}{n^s}-\frac{1}{(n+1)^{s-1}} \right)\frac{\mu(n+1)}{n+1} + \left( \frac{1}{n^s}-\frac{1}{(n+1)^s}\right)\sum_{k=1} ^{n}\frac{\mu(k)}{k}\right].$$ I don't know if it is possible to write it as an interesting identity, I am saying without simplifications – 2017-01-17
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    of your identity when one apply the identity. The second (of those four summands) summand of $\hat{T}(s)$ is $\frac{1}{\zeta(s)}-1$ but as I am saying with this mistakes I don't know if with more calculations one get the identy that you showed or a tautology doing simplification in $1-f(s)=\hat{S}(s)+\hat{T}(s),$ it is I don't know if your function $f(s)$ is simplified and one get only a tautology. Thanks @Agno I am wating an answer, feel free to add your remarks as an answer, that I will read it. – 2017-01-17
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    Thanks for the reference to the method of summation by parts @user1952009 – 2017-01-19

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Numerically it now works, however after simplifying your $S(s)$ and $T(s)$ functions I get:

$$1-f(s)=\sum_{n\ge1}\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right)\cdot\left(\frac{M(n)}{n}+\sum_{k=1}^{n-1}\frac{M(k)}{k(k-1)}-\mu(n+1)-\sum_{m=1}^{n}\frac{\mu(m)}{m}\right)$$

Since $\displaystyle\frac{M(n)}{n}+\sum_{k=1}^{n-1}\frac{M(k)}{k(k-1)}=\sum_{m=1}^{n}\frac{\mu(m)}{m}$ this reduces to:

$$1-f(s)=-\sum_{n\ge1}\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right)\cdot\mu(n+1)$$

which we already had in the previous question, where:

$$f(s)=1+\sum_{n\ge1} (-1)^n \binom{-s}{n}\left(1 - \frac{1}{\zeta(n+s)}\right)$$

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    Then perfect! Many thanks, well my calculations were no goods, but you know now how don't make a mathematical task taking an artificious way and avoid it, currently I am stuck in this kind of confusing reasonings, and will be in next future. Since your identity works for $\Re s>0$ then I consider that your answe solve the question, and I accept it. – 2017-01-17
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    See the [summation by parts](https://en.wikipedia.org/wiki/Summation_by_parts) @user243301 – 2017-01-19