# 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

1. Is it true what I am suspecting for all $$n\in\mathbb N$$?
2. 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?

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}