I am sorry, it's probably very basic question, but I have great hole in head and can't figure it out. I need to separate denominator, like that:
$ \frac { 1 } { n _ { 0 } + n _ { 1 } + n _ { 2 } + \ldots } = \frac { A } { n _ { 0 } } + \frac { B } { n _ { 1 } } + \frac { C } { n _ { 2 } } + \dots $
Please could anyone give some hint. For any help thanks in advance.
There are too many degrees of freedom. This means there are many (infinite even) ways to do this.
For instance, consider the case with two variables: $$ \frac{1}{a+b} = \frac{c}{a} + \frac{d}{b} $$ Notice that you have one equation with two unknown variables! This means we can simply set $c$ to be some constant, e.g. $1$, and then solve for $d$, getting: $$ d = b\left( \frac{1}{a+b} - \frac{c}{a} \right) = b\left( \frac{1}{a+b} - \frac{1}{a} \right) $$ for $c=1$. Of course you could choose anything for $c$.
Let's do it in general. Let: $$ S = \left[ \sum_{i=1}^m x_i \right]^{-1} $$ Then we want to find $\{a_i\}_{i=1}^m$ such that: $$ S = \sum_{i=1}^m \frac{a_i}{x_i} $$ Well, since there is one equation, we have $m-1$ degrees of freedom. Let us fix all the parameters except $a_j$. Essentially we shall freely choose values for $\{a_i\}_{i\ne j}$, at the cost of $a_j$ not being free (i.e. $a_j$ is constrained by the values we choose for $a_i$. Solving for $a_j$ we get: $$ S =\frac{a_j}{x_j} + \sum_{i\ne j} \frac{a_i}{x_i} \;\;\;\implies\;\;\; a_j = x_j\left( S - \sum_{i\ne j} \frac{a_i}{x_i} \right) $$
Cool. So now we can choose some values for the $a_i$'s, and this will fix the value of $a_j$.
For example, let's choose $a_i=x_i\;\forall\;i\ne j$. Then we get: $$ a_k = \begin{cases} x_k & k\ne j\\ x_j(S^{-1}-m+1) & k=j \end{cases} $$
As another example, consider: $ a_i = x_iS/m $ for all $i\ne j$. Then what is $a_j$? Well: $$ a_j=x_j\left( S - \sum_{i\ne j}\frac{S}{m} \right)=x_j\frac{S}{m} $$ So we can say $a_k = x_k S/m$ for all $i\in[1,m]$ for this case.
