Minimum of a function defined by a sum I have the following function, for a fixed integer $n\in \mathbb N$,  $\displaystyle\varphi(x) = \sum_{k=0}^{n} \frac{(-1)^k}{x-k}$, for $x\neq l$ for all $l\in\{0,\ldots,n\}$. I am suspecting that $\min\limits_{x\in (0,n)} \left|\varphi(x)\right|\ge 2$ and want to prove it. So my question is

*

*Is it true what I am suspecting for all $n\in\mathbb N$?

*If not what values of $n$ this is true?

I checked for $n=1$ and I proved easily that the minimum is larger than $4$. Can anyone help me with this in general case?
 A: I think I have found an answer for the even case (when $n$ is even). Indeed, let $n=2p$ and since $\varphi(x) = -\varphi(2p-x)$ it is enough to study the following case:

*

*$x\in(2l-1, 2l)$ for $l\in\{1,\ldots,p\}$
\begin{align}
\left|\varphi(x)\right| &= \left|\sum_{k=0}^{2p} \frac{(-1)^k}{x-k}\right|\\
&= \left|\frac{1}x + \sum\limits_{k=1}^{p} \frac{1}{(x-2k)(x-2k+1)}\right|\\
&= \left|\frac{1}{(x-2l)(x-2l+1)} + \frac1x + \sum\limits_{k=1}^{l-1} \frac{1}{(x-2k)(x-2k+1)} + \sum\limits_{k=l+1}^{p} \frac{1}{(x-2k)(x-2k+1)}\right|\\
&\ge \left|\frac{1}{(x-2l)(x-2l+1)}\right| - \frac1x - \sum\limits_{k=1}^{l-1} \frac{1}{(x-2k)(x-2k+1)} - \sum\limits_{k=l+1}^{p} \frac{1}{(x-2k)(x-2k+1)}\\
&\ge 4 - \frac1{2l-1} - \sum\limits_{k=1}^{l-1} \frac{1}{(2l-1-2k)(2l-1-2k+1)} - \sum\limits_{k=l+1}^{p} \frac{1}{(2l-2k)(2l-2k+1)}\\
&= 4 - \frac1{2l-1}-\sum\limits_{k=1}^{l-1} \frac{1}{(2l-2k)(2l-2k-1)} - \sum\limits_{k=l+1}^{p} \frac{1}{(2l-2k)(2l-2k+1)}\\
&= 4 - \frac1{2l} - \frac{1}{2l(2l-1)}-\sum\limits_{k=1}^{l-1} \frac{1}{2k(2k-1)} - \sum\limits_{k=l+1}^{p} \frac{1}{(2(k-l))(2(k-l)-1)}\\
&\ge 4 - \frac12 - \sum\limits_{k=1}^l \frac{1}{2k(2k-1)} - \sum\limits_{k=1}^{p-l}\\
&\ge 4 - \frac12 - 2\sum_{k=1}^{\infty}\frac{1}{2k(2k-1)}\\
&= 4 - \frac12 - 2\ln 2\ge 2. 
\end{align}
It seems that the odd case can be done by the same method.
