This question is reposted from Math Stack Exchange [(you can see the original post here)][1]. The motivation for reposting is that I feel like the question isn't getting much attention in MSE - if there is something wrong with reposting I have no problem with deleting the question and obviously being sorry for any inconvenience caused. Follows the question:

**Conext and notation.** In the question below, $R_n(z) / S_n(z)$ denotes the n-th convergent of the continued fraction 

$$ \frac{1|}{z-b_1|} + \frac{-a_2|}{z-b_2|} + \dots + \frac{-a_n|}{z-b_n|} + \dots,$$

where $a_n,b_n$ are the coefficients of a generic orthogonal sequence of monic polynomials $\{p_n(z)\}$ that satisfies the three term recurrence relation 

$$ p_n(z) = (x-b_n)p_{n-1}(z) - a_np_{n-2}(z), \quad \text{ for } \, \, n= 1,2,\dots$$

Before this, I have already proved that this n-th convergent can be represented by the formulas

$$ R_n(z) = \int_a^b \frac{p_n(z)-p_n(t)}{z-t}w(t)dt \quad \text{ and } \quad S_n(z) = p_n(z), \quad \text{for } \, \, n =0, 1 , \dots$$

**The problem.** Define the function 

$$ \hat w(z) = \int_a^b \frac{1}{z-t}w(t) dt,$$
which is normally known as the [Stieltjes transformation.][2] I wish to prove that 

$$ \hat w(z) - \frac{R_n(z)}{S_n(z)} = \frac{k_n^1}{z^{2n+1}} + \frac{k_n^2}{z^{2n+2}} + \dots, \quad |z| > R. $$

**My attempt.** I started by using the identity 
$$ \frac{1}{z-t} = \sum_{j=0}^\infty \frac{t^j}{z^{j+1}}, \quad |z| > R, $$
where $R$ is big enough. Since $R$ is big enough, we can guarantee uniform convergence of the series above and then with some auxiliary calculus I was able to obtain the expressions

$$ \hat w(z) = \sum_{j=0}^\infty \frac{\mu_j}{z^{j+1}} \quad \text{ and }  \quad \frac{R_n(z)}{S_n(z)} = \sum_{j=0}^\infty \frac{\mu_j}{z^{j+1}} - \color{blue}{\frac{1}{p_n(z)}\sum_{j=0}^\infty \frac{1}{z^{j+1}} \int_a^b t^j p_n(t) w(t) dt}, $$
where $\mu_j$ represents the j-th moment of the polynomial sequence. Therefore, the difference we want to study is just the blue part in the equation above. It is clear that for $j < n$ we have that the integral in blue is zero (from elementary theory of orthogonal polynomials) and also that for $j=n$ the integral is just $k_n$. So, all we have to do is to study what happens when $j > n.$ Let $j = n+m,$ where $m \in \Bbb N$ is arbitrary. Then,

$$ \int_a^b t^{n+m}p_n(t)w(t)dt = \left[ t^m k_n \right]_a^b - m\int_a^b t^{m-1}k_n dt = 0.$$

Note that what I did above was just to split $t^{n+m} = t^m t^n$ and then applying integration by parts. This tells us that the term in blue is always zero for values of $j$ s.t. $j > n.$ Therefore, the only term in the full infinite summand is just $j=n$ and this yields

$$ \color{blue}{\sum_{j=0}^\infty \frac{1}{z^{j+1}} \int_a^b t^j p_n(t) w(t) dt = \frac{k_n}{z^{n+1}}}. $$

This then guarantees us that 

$$ \hat w(z) - \frac{R_n(z)}{S_n(z)} = \frac{1}{p_n(z)} \frac{k_n}{z^{n+1}},$$

which clearly isn't what I want. This leads me to the conclusion that I have done something wrong but I can't figure out what it was. I am able to give more details about my resolution (I just didn't type step by step to avoid making the post unnecessarily big and hard to read). 

Thanks for any help in advance.


  [1]: https://math.stackexchange.com/questions/4729181/proving-an-identity-related-related-to-stieltjes-transformation-and-continued-fr
  [2]: https://en.wikipedia.org/wiki/Stieltjes_transformation