Approximating $\sum_i \frac{a_i}{\lambda - \lambda_i} = \lambda$ I am trying to solve the equation
$$ \sum_i^n \frac{a_i}{\lambda - \lambda_i}  = \lambda $$
for some real constants $a_i, \lambda_i$, with $\lambda_1 > \lambda_2 > \lambda_3 > ... $ I have the hunch that I can approximate one of its solutions by taking only the first term and solving
$$ \frac{a_1}{\lambda - \lambda_1} = \lambda $$
but I'm having trouble formalizing this step. Under what conditions can I do so and how can I justify this formally? For example, if $\lambda_1 \gg \lambda_2 > \lambda_3 > ...$ then, whenever $\lambda \approx \lambda_1$, the original equation is approximately
$$ \frac{a_1}{\lambda - \lambda_1} + \frac{\sum_{i>1} a_i}{\lambda_1} = \lambda $$
Also, can I find other solutions to the original equation by focusing on other terms? Concretely, what can the solution to the following tell me?
$$ \frac{a_2}{\lambda - \lambda_2} = \lambda. $$
EDIT: I'm especially interested in the case where solving $a_1/(\lambda - \lambda_1) $ is a "good enough" approximation, as computing every $\lambda_1$ might be expensive.
 A: Something I've found to be helpful. We may re-express
\begin{align*}
\sum_{i=1}^{n}\left\{\frac{a_i}{\lambda - \lambda_i} - \frac{\lambda}{n}\right\} = 0
\end{align*}
Solving each summand individually (for positive solutions of $\lambda$):
\begin{align*}
f_i(\lambda) \overset{\text{def}}{=}\frac{a_i}{\lambda - \lambda_i} - \frac{\lambda}{n} = 0 \implies \lambda_i^* = \frac{1}{2}\left(\sqrt{4a_in + \lambda_i^2} + \lambda_i\right)
\end{align*}
Next, approximate $f_i(\lambda)$ with a Taylor series around $\lambda = \lambda_i^*$,
\begin{align*}
f_i(\lambda) &\approx a_i\left\{\frac{1}{\lambda_i^* - \lambda_i} - \frac{\lambda - \lambda_i^*}{\lambda_i^*-\lambda_i} + \frac{(\lambda - \lambda_i^*)^2}{(\lambda_i^*-\lambda_i)^3}\right\} \\
&=A_i\lambda^2 + B_i\lambda + C_i
\end{align*}
for constants $A_i, B_i, C_i$ after collecting the terms. Finally, you then solve
\begin{align*}
0 = \sum_{i=1}^{n}f_i(\lambda) \approx \sum_{i=1}^{n}\{A_i \lambda^2 + B_i \lambda + C_i\} = A\lambda^2 + B\lambda + C
\end{align*}
In reality, Newton-Raphson could get a very close approximation within 4-5 iterations, but with this, you can get a closed-form approximation.
A: Let $f(\lambda) = \sum_i a_i/(\lambda - \lambda_i)$.
Suppose $a_1 > 0$.  Then $\lim_{\lambda \to \lambda_1+} f(\lambda) = +\infty$, so $f(\lambda) > \lambda$ for $\lambda$ near $\lambda_1$ and slightly greater.  On the other hand, $f(\lambda) \to 0$ as $\lambda \to +\infty$, so $f(\lambda) < \lambda$ when $\lambda$ is sufficiently large.  By the Intermediate Value Theorem, there exists a solution to $f(\lambda) = \lambda$ somewhere in the interval $(\lambda_1, +\infty)$.
EDIT: You may be able to use some estimates to show that there is a solution near a solution to $a_1/(\lambda-\lambda_1) = \lambda$.  Write $f(\lambda) = \frac{a_1}{\lambda - \lambda_1} + g(\lambda)$ where
$$ g(\lambda) = \sum_{i=2}^n \frac{a_i}{\lambda - \lambda_i}$$
Suppose $p$ is a solution to $a_1/(\lambda - \lambda_1) = \lambda$, and in some interval $[p-\delta, p+\delta]$ around $p$ we have $f'(\lambda) - 1 < -m < 0$ and 
$|g(\lambda)| < \epsilon$.  The $f(p+\delta) - (p+\delta) < -m \delta + \epsilon$ while
$f(p-\delta) - (p-\delta) > m \delta - \epsilon$.
If $m \delta - \epsilon > 0$, there must be a 
solution to $f(\lambda) = \lambda$ in the interval $(p - \delta, p+\delta)$.  For example, this will work if $a_2,\ldots,a_n$ are sufficiently small.
