TITLE: Some interesting observations on a sum of reciprocals
QUESTION [10 upvotes]: 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?

REPLY [2 votes]: Considering that you look for the first zero of function
$$f(x)=\sum_{i=1}^k \frac 1{x-i}-\frac1 {x-k-1}$$ which can write, using harmonic numbers,
$$f(x)=H_{-x}-H_{k-x}-\frac{1}{x-k-1}$$ remove the asymptotes using
$$g(x)=(x-1)(x-2)f(x)=2x-3+(x-1)(x-2)\left(H_{2-x}-H_{k-x}-\frac{1}{x-k-1} \right)$$ You can approximate the solution using a Taylor expansion around $x=1$ and get
$$g(x)=-1+(x-1) \left(-\frac{1}{k}+\psi ^{(0)}(k)+\gamma
   +1\right)+O\left((x-1)^2\right)$$ Ignoring the higher order terms, this gives as an approximation
$$x_{est}=1+\frac{k}{k\left(\gamma  +1+ \psi ^{(0)}(k)\right)-1}$$ which seems to be "decent"  (and, for sure, confirms your claims).
$$\left(
\begin{array}{ccc}
k & x_{est} &  x_{sol} \\
 2  & 1.66667 & 1.58579 \\
 3  & 1.46154 & 1.46791 \\
 4  & 1.38710 & 1.41082 \\
 5  & 1.34682 & 1.37605 \\
 6  & 1.32086 & 1.35209 \\
 7  & 1.30238 & 1.33430 \\
 8  & 1.28836 & 1.32040 \\
 9  & 1.27726 & 1.30914 \\
 10 & 1.26817 & 1.29976 \\
 11 & 1.26055 & 1.29179 \\
 12 & 1.25403 & 1.28489 \\
 13 & 1.24837 & 1.27884 \\
 14 & 1.24339 & 1.27347 \\
 15 & 1.23895 & 1.26867 \\
 16 & 1.23498 & 1.26433 \\
 17 & 1.23138 & 1.26039 \\
 18 & 1.22810 & 1.25678 \\
 19 & 1.22510 & 1.25346 \\
 20 & 1.22233 & 1.25039
\end{array}
\right)$$ For infinitely large values of $k$, the asymptotics of the estimate would be
$$x_{est}=1+\frac{1}{\log \left({k}\right)+\gamma +1}$$
For $k=1000$, the exact solution is $1.12955$ while the first approximation gives $1.11788$ and the second $1.11786$.
Using such estimates would make Newton method converging quite fast (shown below for $k=1000$).
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 1.117855442 \\
 1 & 1.129429575 \\
 2 & 1.129545489 \\
 3 & 1.129545500
\end{array}
\right)$$
Edit
We can obtain much better approximations if, instead of using a Taylor expansion of $g(x)$ to $O\left((x-1)^2\right)$, we build the simplest $[1,1]$ Padé approximant (which is equivalent to an  $O\left((x-1)^3\right)$ Taylor expansion). This would lead to
$$x=1+ \frac{6 (k+k (\psi ^{(0)}(k)+\gamma )-1)}{\pi ^2 k+6 (k+\gamma  (\gamma 
   k+k-2)-1)-6 k \psi ^{(1)}(k)+6 \psi ^{(0)}(k) (2 \gamma  k+k+k \psi
   ^{(0)}(k)-2)}$$ Repeating the same calculations as above, the results are
$$\left(
\begin{array}{ccc}
k & x_{est} &  x_{sol} \\
 2  & 1.60000 & 1.58579 \\
 3  & 1.46429 & 1.46791 \\
 4  & 1.40435 & 1.41082 \\
 5  & 1.36900 & 1.37605 \\
 6  & 1.34504 & 1.35209 \\
 7  & 1.32741 & 1.33430 \\
 8  & 1.31371 & 1.32040 \\
 9  & 1.30266 & 1.30914 \\
 10 & 1.29348 & 1.29976 \\
 11 & 1.28569 & 1.29179 \\
 12 & 1.27897 & 1.28489 \\
 13 & 1.27308 & 1.27884 \\
 14 & 1.26787 & 1.27347 \\
 15 & 1.26320 & 1.26867 \\
 16 & 1.25899 & 1.26433 \\
 17 & 1.25516 & 1.26039 \\
 18 & 1.25166 & 1.25678 \\
 19 & 1.24844 & 1.25346 \\
 20 & 1.24547 & 1.25039
\end{array}
\right)$$
For $k=1000$, this would give as an estimate $1.12829$ for an exact value of  $1.12955$.
For infinitely large values of $k$, the asymptotics of the estimate would be
$$x_{est}=1+\frac{6 (\log (k)+\gamma +1)}{6 \log (k) (\log (k)+2 \gamma +1)+\pi ^2+6 \gamma 
   (1+\gamma )+6}$$