## Abstract

In addition to other answers I'd like to give a more analytically centered approach. I'll show that for $\operatorname{Re}(x')>0.$ it holds that $-(\psi^{(1)}(x'+n)-\psi^{(1)}(x'))=\sum_{i=1}^{n}\frac{1}{(i-1+x')^2}$. With that we can show why mathematica is "generically right" but wrong in the specific case considered here.

## Proof

According to [wikipedia][1] it holds that 

$$
-[\psi(x'+n)-\psi(x')]=-\sum_{i=0}^{n-1}\frac{1}{i+x'}
$$

for $\operatorname{Re}(x')>0$. If we now take the derivative of left and right hand side, we obtain the polygamma functions found in the mathemtica expression on the left hand side.

$$
-[\psi^{(1)}(x'+n)-\psi^{(1)}(x')]=\sum_{i=0}^{n-1}\frac{1}{(i+x')^2}
$$

For convenience we shift the expression on the right hand side a little and obtain the desired result.

$$
-[\psi^{(1)}(x'+n)-\psi^{(1)}(x')]=\sum_{i=1}^{n}\frac{1}{(i-1+x')^2}
$$

## Applying the result to the specific case

We set $x'=(1-2n)$ in the upper formula. This clearly violates the premise, because `x'` will be negative. However, we will proceed by evaluating the expressions formally to explain mathematicas result.

$$
\begin{split}
-[\psi^{(1)}((1-n)-\psi^{(1)}(1-2n)]&=\psi^{(1)}(1-2n)-\psi^{(1)}((1-n)\\
&=\sum_{i=1}^{n}\frac{1}{(i-1+(1-2n))^2}
\end{split}
$$

The left hand side equals mathematica `Polygammes[1,1-2n]-Polygamma[1,1-n]` and the denominator of the sum equals `(i + (n - i) x)^2 /. x -> 2`. This can be checked with

    (i - 1 + (1 - 2 n))^2 == (i + (n - i)*x)^2 /. x -> 2 // FullSimplify

which gives `True`.

To stress my point again $x'=-1+(1-2 n)$ is negative for any $n\geq0$. Since mathematica cannot know what `x` is going to be in advance I assume that it just uses some generic transformations, which do not apply in all cases and especially not in this case.

** **Edit 1** **

I'm not quite sure if the strict requirement $\operatorname{Re}(x')>0$ is really necessary for the transformation to hold or if the weaker requirement $x'\neq0,-1,-2,-3,...$ is sufficient. The term $\psi^{(1)}(1-2n)-\psi^{(1)}((1-n)$ is nevertheless divergent at least for $n\in\mathbb{N}>0$. This means the result is indeed wrong.

** **Edit 2** **

In this edit I will elaborate on the limiting procedure. I will suggest a limiting procedure different from the limiting procedure `Limit[PolyGamma[1,1-2n] - PolyGamma[1,1-n], n -> 10]` in the OP. The limit considered here, namely `Limit[PolyGamma[1,1-z] - PolyGamma[1,1-z+n], z ->2n]` with `n=10`, yields the same result as `Sum[ 1/(i + (n - i) x)^2, {i, 1, n}]` with `n=10` and `x=2`. Furthermore, I will also explain why I deem the limiting procedure suggested in OP as unnatural.

## Limit

I assume that mathematica analyzes the expression inside the sum, possibly by pattern matching, and then applies some generic transformation. The sum can also be written as

$$ \tag{1}\label{sum-equation}
\sum_{j=1}^{n}\frac{1}{(i+(n-i)x)^2}=\sum_{j=1}^{n}\frac{1}{(ia+z)^2}
$$
with $a=1-x$ and $z=nx$.By evaluating the right hand side `Sum[1/(j*a + z)^2, {j,1,n}]`, we obtain

$$\tag{2}\label{sum-polygamma-relation}
\sum_{j=1}^{n}\frac{1}{ia+z}=\frac{1}{a^2}\left(\psi^{(1)}(1+z/a)-\psi^{(1)}(1+z/a+n)\right)
$$
I will now take the right hand side and treat $z$ as the idependent variable rather than $n$. The left hand side has poles at $z=-\frac{1}{a},-\frac{2}{a},...,-\frac{n}{a}$. The right hand side has those poles as well, but additionally it has points where it is not defined. Those points are at $z=-\frac{j}{n}$ with $j\in\mathbb{N}>n$. Now if we want to visualize that we pick `x=2` and `n=10` this gives us `a=-1` and `z=-20`. We can now plot either the `righthandside` or `lefthandside`

    lefthandside = Sum[1/(-i + z)^2, {i, 1, n}];
    righthandside = PolyGamma[1, 1 - z] - PolyGamma[1, 1 - z + n];
    Show[Plot[lefthandside, {z, -22, 15}, PlotRange -> {0, 15}, AspectRatio -> 1], Graphics[{Red, Circle[{-20, 0.0539}, 0.5]}], PlotRangePadding -> {{0, 0}, {1.5, 0}}, AspectRatio -> Automatic]

[![plot-sum-polygamma][2]][2]

and arrive at an identical plot for both. The red circle marks left- and right hand sides value at `z=-20` which is about `0.0539`. Note that the right hand side of \eqref{sum-polygamma-relation} is not defined at $z=-1,-2,-3,...$. Nevertheless it becomes obvious from the plot that there exists a continuous extension to all of $\mathbb{C}$. This continuous extension is given by the sum. When excluding the singular points left and right hand side agree on all of $\mathbb{C}$.  

## Conclusion

IMHO considering $n$ as the independent variable in the limiting procedure seems unnatural as non-integer, negativ or complex values for $n$ do not make sense as the upper limit of the sum. Hence it seems more natural to rewrite the sum in terms of polygamma functions as suggested in \eqref{sum-polygamma-relation} and evaluate those in a neighborhood of $z=xn$. For $x\not\in\mathbb{N}$ the polygamma functions given by mathematica agree with the sum. For $x\in\mathbb{N}$ the right hand side of \eqref{sum-polygamma-relation} is ill defined. Contrary to my previous assertion in "edit 1" the polygamma term does not diverge there. It is just not defined. Since the expression given by mathematica is only wrong on a subset of $\mathbb{C}$ one may or may not deem the mathemtica solution as "generically" correct.


  [1]: https://en.wikipedia.org/wiki/Digamma_function#Recurrence_formula_and_characterization
  [2]: https://i.sstatic.net/W59vB.gif