A Really Basic Infinite Sum Problem (This might also be a geometric series)

Question

I'm not really sure how I should go about this problem:

$\displaystyle \frac{x-1}{x} +\frac{x-2}{x} +\frac{x-3}{x} \ .\ .\ .\ +\frac{1}{x} =3,\ x\in ℕ$

I tried multiplying by x on both sides, but this just results in 3 being equal to infinity times x.

Answer 1

Hint:

The sum of the numerators is $\dfrac {x(x-1)}2$.

Answer 2

Note that\begin{align}\frac{x-1}x+\frac{x-2}x+\cdots+\frac1x&=\left(1-\frac1x\right)+\left(1-\frac2x\right)+\cdots+\left(1-\frac{x-1}x\right)\\&=x-1-\frac1x\bigl(1+2+\cdots+(x-1)\bigr)\\&=x-1-\frac1x\times\frac{x(x-1)}2\\&=\frac{x-1}2.\end{align}Can you take it from here?