Sum of integer reciprocals

You can do just a little bit of trial and error. If the summands are $1/n, 1/(n + 1)$ and $1/(n + 2)$, then we ought to have

$$\frac 1 {n + 1} \approx \frac 1 3 \cdot \frac{47}{60} \approx \frac 1 4$$

because $1/(n + 1)$ is reasonably close to the average of the three reciprocals. And it turns out that

$$\frac 1 3 + \frac 1 4 + \frac 1 5 = \frac{47}{60}.$$

we have $$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}=\frac{47}{60}$$ this equation is equivalent to $${\frac { \left( n-3 \right) \left( 47\,{n}^{2}+102\,n+40 \right) }{ 60\,n \left( n+1 \right) \left( n+2 \right) }} =0$$ can you finish now?

This answer is based upon symmetry and a guess.

We try to exploit symmetry and use the approach \begin{align*} \color{blue}{\frac{47}{60}}&=\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}\\ &=\frac{n(n+1)+n^2-1+(n-1)n}{n(n^2-1)}\\ &\color{blue}{=\frac{3n^2-1}{n(n^2-1)}} \end{align*}

Let's take a closer look at \begin{align*} \frac{3n^2-1}{n(n^2-1)}=\frac{47}{60} \end{align*}

Hoping for a simple proportion we equate the numerators and get \begin{align*} 3n^2-1=47\quad\Rightarrow\quad 3n^2=48\quad\Rightarrow\quad n\in\{4,-4\} \end{align*}

The denominator $n(n^2-1)$ evaluated at $n=4$ gives $4(16-1)=60$ and we get the solution $$\color{blue}{n=4}$$

Note: Observe, it was the symmetry which enabled us to make the lucky guess.