Use Vieta!

For this problem, we will relate $a,b,c$ to the roots of some easy to factor polynomial, making the sum a breeze!

First, let's consider the second equation given:

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{k} \Rightarrow abc = k(ab+bc+ac)$

Let $ab+bc+ac=m$ . Then $abc=mk$ .

These equations look an awful lot like Vieta's Formula! Consider the following cubic:

$x^{3}-kx^{2}+mx-mk=0$

Based on our conditions, this polynomial has roots $a,b,c$ ! We can factor it as follows:

$x^{2}(x-k)+m(x-k)=0 \Rightarrow (x-\sqrt{m}i)(x+\sqrt{m}i)(x-k)=0$

Where $i = \sqrt{-1}$ . Therefore $a,b,c = \sqrt{m}i,-\sqrt{m}i,k$

Since our sum is adding $a,b,c$ raised to odd powers, the $\sqrt{m}i$ and $-\sqrt{m}i$ terms will cancel for every $j$ in out summand. This makes the sum:

$\sum_{j=0}^{\infty}k^{2j+1} = k+k^{3}+k^{5}+k^{7}+ \ldots$

Which is an infinite geometric series with initial term $k$ and ratio $k^{2}$ , meaning that if $|k|<1$ , then the sum is $\boxed{\frac{k}{1-k^{2}}}$

Note that we could have $m<0$ , in which case our polynomial would factor as $(x+\sqrt{m})(x-\sqrt{m})(x-k)=0$ , but adding the roots to odd powers would still have the $\pm$ terms cancel

$\color{#3D99F6} \sum_{j=0}^{\infty} a^{2j+1}+b^{2j+1}+c^{2j+1}$

This is a GP of course. The sum of the above GP is = $\color{#D61F06} \large \dfrac{a}{1-a^{2}}+ \dfrac{b}{1-b^{2}}+ \dfrac{c}{1-c^{2}}$

Now we also get $\color{#E81990} \large \dfrac{1}{a+b+c}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$

And we get $\color{#20A900} \large \dfrac{1}{a+b+c}-\dfrac{1}{c}= \dfrac{1}{a}+\dfrac{1}{b}$ The above expression when simplified results as

$\color{#CEBB00} \large \dfrac{-(a+b)}{(a+b+c)c}= \dfrac{a+b}{ab}$ this results in

$\color{grey} \large ab=-ac-bc-c^{2}$ = $\color{grey} \large a(b+c)=-c(b+c)$ This gives us $\color{grey} \large a=-c$ and this results in

$\color{cyan} \huge \boxed{\boxed{\boxed{b=k}}}$ and $\color{cyan} \dfrac{b}{1-b^{2}} = \dfrac{k}{1-k^{2}}$ .