Solve the equation $\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x+4}=0$ Solve the equation $$\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0.$$
For $x\ne -4;-3;-2;-1;0$ we have $$(x+1)(x+2)(x+3)(x+4)+x(x+2)(x+3)(x+4)+x(x+1)(x+3)(x+4)+\text{...}=0$$
Most likely that's not the author's intention. I have tried to substitute $t=x+2$ to get $$\dfrac{1}{t-2}+\dfrac{1}{t-1}+\dfrac{1}{t}+\dfrac{1}{t+1}+\dfrac{1}{t+2}=0$$ which actually isn't easier to work with than the original problem.

Let $f$ be the conveniently translated function defined by :
$$f(t):=\dfrac{1}{t-2}+\dfrac{1}{t-1}+\dfrac{1}{t}+\dfrac{1}{t+1}+\dfrac{1}{t+2}$$
What I would like to underline is that a graphical representation provides a good insight for such questions. In particular, it can help to anticipate (or verify)  a certain number of results, in particular the fact that the roots of $f(t)=0$ are all real, separated by integers $-2,-1,0,1,2$ and symmetric with respect to $0$ due to the fact that $f$ is an odd function.

Indeed, this function has a decreasing behavior (the derivative $f'$ is clearly negative) from $+\infty$ to $+\infty$ on each of the intervals: $[k,k+1], \ k=-2,-1,0,1$ delimitated by vertical asymptotes ; the two external branches do not provide any supplementary root.
