Upper bound on summation involving fractional part Let $x\in[1,+\infty)\subset\mathbb{R}$. I would like to show that
$$\sum_{d=1}^{\lfloor x\rfloor}\left(\frac{x}{d}-\left\lfloor\frac{x}{d}\right\rfloor\right)\leq x-1,$$
where $\lfloor\cdot\rfloor$ is the floor function. I understand that, since $$\frac{x}{d}-\left\lfloor\frac{x}{d}\right\rfloor\leq1,$$
we have
$$\sum_{d=1}^{\lfloor x\rfloor}\left(\frac{x}{d}-\left\lfloor\frac{x}{d}\right\rfloor\right)\leq \sum_{d=1}^{\lfloor x\rfloor}1=\lfloor x\rfloor\leq x.$$
How can I come up with the finer bound? Any hints? Thank you in advance.
 A: Your sum can be re-written as 
$$
\sum_{d \leq x} \Big ( \frac{x}{d} - \sum_{\substack{n \leq x \\ d | n}} 1 \Big )
$$
Interchanging the two sums you get
$$
x \sum_{d \leq x} \frac{1}{d} - \sum_{n \leq x} d(n)
$$
where $d(n)$ counts the number of divisors of $n$. Thus the problem is reducing to computing
$$
\sum_{n \leq x} d(n).
$$
There is a geometric interpretation of this sum as counting the number of lattice points inside the hyperbola $x y = n$. Using this geometric interpretation (the so-called "Hyperbola method" of Dirichlet) you can show that, 
$$
\sum_{n \leq x} d(n) = x \log x + (2 \gamma - 1) x + O(\sqrt{x})
$$
where $\gamma$ is the Euler-Mascheroni constant.
Recall that
$$
\sum_{d \leq x} \frac{1}{d} = \log x + \gamma + O(1/x).
$$
So this gives,
$$
x \sum_{d \leq x} \frac{1}{d} - \sum_{n \leq x} d(n) = (1 - \gamma) x + O(\sqrt{x})
$$
and therefore proves your claim (in stronger form) for all large $x$. Working out explicit constants in $O(\cdot)$ (which is easy by looking it up on the internet) will give you a bound and you can check on a computer that the claim is true for the remaining values. Note that this problem is a bit delicate because you cannot beat $x$ by too much, the asymptotic is actually $(1 - \gamma) x$ and $1 - \gamma \approx 0.4227843351$.
A: We have
\begin{align*}
& \sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\left( {\frac{x}{d} - \left\lfloor {\frac{x}{d}} \right\rfloor } \right)}  = \sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\left( {\frac{x}{d} - \left\lfloor {\frac{{\left\lfloor x \right\rfloor }}{d}} \right\rfloor } \right)}  = \sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\left( {\frac{x}{d} - \frac{{\left\lfloor x \right\rfloor }}{d}} \right)}  + \sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\left( {\frac{{\left\lfloor x \right\rfloor }}{d} - \left\lfloor {\frac{{\left\lfloor x \right\rfloor }}{d}} \right\rfloor } \right)} 
\\ & \le (x - \left\lfloor x \right\rfloor )\sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\frac{1}{d}}  + \sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\left( {1 - \frac{1}{d}} \right)}  = (x - \left\lfloor x \right\rfloor  - 1)\sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\frac{1}{d}}  + \sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } 1 
\\ & = x - 1 + (x - \left\lfloor x \right\rfloor  - 1)\sum\limits_{d = 2}^{\left\lfloor x \right\rfloor } {\frac{1}{d}}  \le x - 1.
\end{align*}
