Some interesting observations on a sum of reciprocals This recent question is the motivation for this post.


Consider the following equation $$\frac1{x-1}+\frac1{x-2}+\cdots+\frac1{x-k}=\frac1{x-k-1}$$ where $k>1$.
My claims:

*

*There are $k$ solutions, all of which are real.


*Let $x_{\min}$ be the minimum value of these $k$ solutions. Then as $k\to\infty$, $x_{\min}$ converges. (If it does, to what value does it converge?)


*As $k\to\infty$, all of the solutions get closer and closer to an integer, which is bounded below. Furthermore, these integers will be $1, 2, 3, \cdots, k-1, k+1$.

To see these patterns, I provide the solutions of $x$ below. I used W|A for $k\ge4$. The values in $\color{blue}{\text{blue}}$ are those of $x_{\min}$.
$$\begin{array}{c|c}k&2&3&4&5&6\\\hline x&4.414&4.879&5.691&6.592&7.530\\&\color{blue}{1.585}&2.652&3.686&4.701&5.722\\&&\color{blue}{1.468}&2.545&3.588&4.615\\&&&\color{blue}{1.411}&2.487&3.531\\&&&&\color{blue}{1.376}&2.449\\&&&&&\color{blue}{1.352}\end{array}$$
Also, when $k=2$, the polynomial in question is $x^2-6x+7$, and when $k=3$, it is $x^3-9x^2+24x-19$.
The reason why I think $x_{\min}$ converges is because the difference between the current one and the previous gets smaller and smaller as $k$ increases.


Are my claims true?

 A: For the base case,
$$\tag1f_2(x)=\frac1{x-1}+\frac1{x-2}-\frac1{x-3}, $$
one readily verifies that there is a root in $(1,2)$ and a root $x^*$ in $(3,+\infty)$.
If we multiply out the denominators of
$$f_k(x)=\frac1{x-1}+\frac1{x-2}+\ldots+\frac1{x-k}-\frac1{x-k-1},$$
we obtain the equation 
$$\tag2(x-1)(x-2)\cdots(x-k-1)f_k(x)=0,$$
which is a polynomial of degree (at most) $k$, so we expect $k$ solutions, but some of these may be complex or repeated or happen to be among $\{1,2,\ldots, k+1\}$ and thus not allowed for the original equation.
But $f_k(x)$
has simple poles with jumps from $-\infty$ to $+\infty$ at $1,2,3,\ldots, k$, and a simple pole with jump from $+\infty$ to $-\infty$ at $k+1$, and is continuous otherwise. It follows that there is (at least) one real root in $(1,2)$, at least one in in $(2,3)$, etc. up to $(k-1,k)$, so there are at least $k-1$ distinct real roots.
Additionally, for $x>k+1$ and $k\ge2$, we have
$$f_k(x)\ge f_2(x+k-2).$$
It follows that there is another real root between $k+1$ and $x^*+k-2$.
So indeed, we have $k$ distinct real roots.
From the aboive, the smallest root is always in $(1,2)$.
If follows from $f_{k+1}(x)>f_k(x)$ for $x\in(1,2)$ and the fact that all $f_k$ are strictly decreasing there, that $x_\min $ decreases with increasing $k$. As a decreasing bounded sequence, it does have a limit.
A: Both your claims are true.
if you call 
$$
f(x) = \frac1{x-1}+\frac1{x-2}+\cdots+\frac1{x-k}-\frac1{x-k-1}
$$
then $f(1^+) = +\infty$, $f(2^-) = -\infty$ and $f$ is continuous in $(1,2)$, so it has a root in $(1,2)$. 
The same you can say about $(2,3)$, $(3,4), \cdots, (k-1,k)$, so there are at least $k-1$ real distinct roots. $f$ is also equivalent to a $k$-degree polynomial with the same root, but a $k$-degree polynomial with $k-1$ real roots has in reality $k$ real roots.
The last root lies in $(k+1,+\infty)$, since $f(k+1^+) = -\infty$ and $f(+\infty) = +\infty$. 
The least root $x_{\min}$ must lie in $(1,2)$, since $f(x)<0$ for every $x<1$. Moreover, 
$$
f(x) = 0\implies x = 1 + \frac{1}{\frac1{x-k-1}-\frac1{x-2}-\cdots-\frac1{x-k}}
$$
and knowing $1<x<2$, we infer $\frac1{x-k-1}>\frac1{x-2}$ and
$$
1<x = 1 + \frac{1}{\frac1{x-k-1}-\frac1{x-2}-\cdots-\frac1{x-k}}
< 1 - \frac{1}{\frac1{x-3}+\cdots+\frac1{x-k}}\to 1
$$
so $x_{\min}$ converges to $1$

About the third claim, notice that you may repeat the same argument for any root except the biggest. Let us say that $x_r$ is the $r-th$ root, with $r<k$, and we know that $r<x_r<r+1$. 
$$
f(x_r) = 0\implies x_r = r + \frac{1}{\frac1{x_r-k-1}-\frac1{x_r-1}-\cdots-\frac1{x_r-k}}
$$
but $\frac1{x_r-k-1}>\frac1{x_r-1}$  holds, so
$$
r<x_r = r + \frac{1}{\frac1{x_r-k-1}-\frac1{x_r-1}-\cdots-\frac1{x_r-k}}
< r - \frac{1}{\frac1{x_r-2}+\cdots+\frac1{x_r-k}}\to r
$$
so $x_r$ converges to $r$.
For the biggest root, we know $k+1<x_k$ and
$$
f(x_k) = 0\implies k+1 < x_k = k+1 + \frac{1}{\frac1{x_k-1}+\cdots+\frac1{x_k-k}} \to k+1
$$
