Is there a direct way of proving that $$ [nx] = [x] + [x+\frac{1}{2}] + [x+\frac{1}{3}] + \ldots + [x+ \frac{1}{n}]$$ for each real number $x$ and for each positive integer $n$?
My effort:
Let $n$ be an arbitrary but fixed positive integer. Then since we can write $$x = [x]+ \theta,$$ where $0 \leq \theta < 1$, we can consider the following cases:
Case 1. When $0 \leq \theta < 1/n$, we see that $$ nx = n [x] + n\theta$$ and $0 \leq n \theta < 1$; so we have $$[nx] = n[x].$$ Also, for $n \geq 2$, we have $$ \frac{1}{2} \leq \theta + \frac{1}{2} < \frac{1}{n} + \frac{1}{2} \leq 1, $$ whence $$ x + \frac{1}{2} \leq [x] + \theta + \frac{1}{2}$$ so that $$[x+\frac{1}{2}] = [x].$$ Continuing in the same way, we find that $$ \frac{1}{n} \leq \theta + \frac{1}{n} < \frac{1}{n} + \frac{1}{n} = \frac{2}{n} \leq 1$$ whenever $n \geq 2$, which implise that $$ x + \frac{1}{n} \leq [x] + \theta + \frac{1}{n} $$ so that $$[x + \frac{1}{n}] = [x]. $$ Finally, adding all these results, we find that $$ [x] + [x+\frac{1}{2}] + [x+\frac{1}{3}] + \ldots + [x+ \frac{1}{n}] = n[x] = [nx], $$ as required.
Case 2. When $\frac{1}{n} \leq \theta < \frac{2}{n}$, we have $1 \leq n \theta < 2$, and since $ nx = n[x] + n\theta$, we can conclude that $$[nx] = n[x] + 1.$$ And then we handle the right-hand side as in case 1 above.
Is there a more efficient, but elementary, way of deriving the above formula?
There is no direct way of proving your claim because it is wrong. Let $n=3$ and $x=\frac25$.