[proofplan] We count the same set of lattice points in two ways. The sum $\sum_{1 \leq n \leq x}\tau(n)$ counts ordered pairs $(a,b) \in \mathbb{N}^2$ with $ab \leq x$. Dirichlet's hyperbola method reduces this count to a sum over $a \leq \sqrt{x}$, where replacing floors by their main terms introduces only $O(\sqrt{x})$ error. The harmonic-number asymptotic then gives the constant term $2\gamma - 1$. [/proofplan] [step:Rewrite the divisor sum as a lattice point count below the hyperbola] Define the divisor summatory function \begin{align*} S: [1, \infty) &\to \mathbb{R} \\ x &\mapsto \sum_{1 \leq n \leq x}\tau(n). \end{align*} Since $\tau(n)$ counts the positive divisors of $n$, we have \begin{align*} S(x) &= \#\{(d,m) \in \mathbb{N}^2 : dm \leq x\}. \end{align*} Indeed, the map \begin{align*} (d,m) \mapsto dm \end{align*} identifies each ordered pair with a divisor $d$ of the integer $n = dm$, and every divisor $d \mid n$ gives the ordered pair $(d,n/d)$. [/step] [step:Apply the hyperbola decomposition at $\sqrt{x}$] Let \begin{align*} y := \lfloor \sqrt{x}\rfloor. \end{align*} Every pair $(d,m) \in \mathbb{N}^2$ with $dm \leq x$ has either $d \leq y$ or $m \leq y$, unless both $d > y$ and $m > y$. But if $d > y$ and $m > y$, then $d,m \geq y+1 > \sqrt{x}$, so $dm > x$, which is impossible. Therefore the union \begin{align*} A_x &:= \{(d,m) \in \mathbb{N}^2 : dm \leq x,\ d \leq y\},\\ B_x &:= \{(d,m) \in \mathbb{N}^2 : dm \leq x,\ m \leq y\} \end{align*} equals the full set $\{(d,m) \in \mathbb{N}^2 : dm \leq x\}$. By symmetry, \begin{align*} \#A_x = \#B_x = \sum_{1 \leq d \leq y}\left\lfloor \frac{x}{d}\right\rfloor. \end{align*} Their intersection is \begin{align*} A_x \cap B_x = \{(d,m) \in \mathbb{N}^2 : d \leq y,\ m \leq y\}, \end{align*} because $dm \leq y^2 \leq x$ whenever $d,m \leq y$. Hence $\#(A_x \cap B_x)=y^2$, and inclusion-exclusion gives \begin{align*} S(x) = 2\sum_{1 \leq d \leq y}\left\lfloor \frac{x}{d}\right\rfloor - y^2. \end{align*} [guided] The divisor sum counts all integer lattice points below the hyperbola $dm=x$ in the first quadrant. The hyperbola method counts the part with first coordinate at most $\sqrt{x}$ and the symmetric part with second coordinate at most $\sqrt{x}$. Set \begin{align*} y := \lfloor \sqrt{x}\rfloor. \end{align*} Define \begin{align*} A_x &:= \{(d,m) \in \mathbb{N}^2 : dm \leq x,\ d \leq y\},\\ B_x &:= \{(d,m) \in \mathbb{N}^2 : dm \leq x,\ m \leq y\}. \end{align*} If a pair $(d,m)$ satisfies $dm \leq x$ and is not in $A_x \cup B_x$, then $d>y$ and $m>y$. Since $d,m$ are integers, this implies $d,m \geq y+1$. Because $y=\lfloor \sqrt{x}\rfloor$, we have $y+1>\sqrt{x}$, and therefore \begin{align*} dm \geq (y+1)^2 > x, \end{align*} contradicting $dm \leq x$. Thus $A_x \cup B_x$ is exactly the set being counted. For fixed $d \leq y$, the possible values of $m$ are precisely \begin{align*} 1 \leq m \leq \frac{x}{d}, \end{align*} so there are $\left\lfloor x/d\right\rfloor$ such integers $m$. Hence \begin{align*} \#A_x = \sum_{1 \leq d \leq y}\left\lfloor \frac{x}{d}\right\rfloor. \end{align*} By the symmetry exchanging $d$ and $m$, the same formula holds for $\#B_x$. The overlap consists of pairs with both coordinates at most $y$. For such pairs, \begin{align*} dm \leq y^2 \leq x, \end{align*} so the hyperbola condition is automatic. Hence \begin{align*} A_x \cap B_x = \{(d,m) \in \mathbb{N}^2 : d \leq y,\ m \leq y\}, \end{align*} and $\#(A_x \cap B_x)=y^2$. Inclusion-exclusion gives \begin{align*} S(x) = \#A_x+\#B_x-\#(A_x\cap B_x) = 2\sum_{1 \leq d \leq y}\left\lfloor \frac{x}{d}\right\rfloor - y^2. \end{align*} [/guided] [/step] [step:Replace the floor sum by the harmonic main term] Let \begin{align*} H_y := \sum_{1 \leq d \leq y}\frac{1}{d} \end{align*} denote the $y$-th harmonic number. In this proof, the notation $R(x)=O(G(x))$ as $x \to \infty$ means that there exist constants $C>0$ and $x_0 \geq 1$ such that $|R(x)| \leq C|G(x)|$ for all $x \geq x_0$. For each $d \in \mathbb{N}$, \begin{align*} \frac{x}{d} - 1 < \left\lfloor \frac{x}{d}\right\rfloor \leq \frac{x}{d}. \end{align*} Summing this inequality over $1 \leq d \leq y$ gives \begin{align*} \sum_{1 \leq d \leq y}\left\lfloor \frac{x}{d}\right\rfloor = xH_y + O(y), \end{align*} where the implicit constant is absolute. Therefore \begin{align*} S(x)=2xH_y-y^2+O(y). \end{align*} Since $y=\lfloor \sqrt{x}\rfloor$, we have $y=O(\sqrt{x})$, and hence $O(y)=o(x)$. [guided] The floor function is the only source of arithmetic irregularity in the hyperbola formula. We isolate it by comparing each floor with its real value. We use the notation $R(x)=O(G(x))$ as $x \to \infty$ to mean that there exist constants $C>0$ and $x_0 \geq 1$ such that $|R(x)| \leq C|G(x)|$ for all $x \geq x_0$. For every integer $d \geq 1$, \begin{align*} \frac{x}{d} - 1 < \left\lfloor \frac{x}{d}\right\rfloor \leq \frac{x}{d}. \end{align*} Define the harmonic number \begin{align*} H_y := \sum_{1 \leq d \leq y}\frac{1}{d}. \end{align*} Adding the floor inequalities for $1 \leq d \leq y$ yields \begin{align*} xH_y - y < \sum_{1 \leq d \leq y}\left\lfloor \frac{x}{d}\right\rfloor \leq xH_y. \end{align*} Equivalently, \begin{align*} \sum_{1 \leq d \leq y}\left\lfloor \frac{x}{d}\right\rfloor = xH_y + O(y), \end{align*} with an absolute implied constant. Substituting this into the hyperbola identity gives \begin{align*} S(x) =2\sum_{1 \leq d \leq y}\left\lfloor \frac{x}{d}\right\rfloor-y^2 =2xH_y-y^2+O(y). \end{align*} Because $y=\lfloor \sqrt{x}\rfloor$, the error term satisfies $O(y)=O(\sqrt{x})=o(x)$. Thus the only remaining task is to evaluate $2xH_y-y^2$ to precision $o(x)$. [/guided] [/step] [step:Evaluate the harmonic number at $y=\lfloor\sqrt{x}\rfloor$] By the definition of Euler's constant, \begin{align*} H_y = \log y + \gamma + o(1) \end{align*} as $y \to \infty$. Since $y=\lfloor \sqrt{x}\rfloor$, we have $y \to \infty$ as $x \to \infty$, and hence \begin{align*} 2xH_y = 2x\log y + 2\gamma x + o(x). \end{align*} Moreover, \begin{align*} 0 \leq \sqrt{x}-y < 1, \end{align*} so \begin{align*} y^2 = x + O(\sqrt{x}) = x + o(x). \end{align*} Also $y/\sqrt{x} \to 1$, so \begin{align*} 2\log y = \log x + 2\log\left(\frac{y}{\sqrt{x}}\right) = \log x + o(1). \end{align*} Multiplying by $x$ gives \begin{align*} 2x\log y = x\log x + o(x). \end{align*} Therefore \begin{align*} 2xH_y-y^2 = x\log x + (2\gamma-1)x + o(x). \end{align*} [/step] [step:Combine the estimates to obtain the divisor summatory asymptotic] From the previous steps, \begin{align*} S(x) =2xH_y-y^2+O(y) \end{align*} with $y=\lfloor \sqrt{x}\rfloor$. Since $O(y)=o(x)$ and \begin{align*} 2xH_y-y^2 = x\log x + (2\gamma-1)x + o(x), \end{align*} we obtain \begin{align*} \sum_{1 \leq n \leq x}\tau(n) = S(x) = x\log x + (2\gamma-1)x + o(x). \end{align*} This is the claimed asymptotic as $x \to \infty$. [/step]