<s>[INST] Sum of number-of-divisors function equals $\sum_{j=1}^{n} \lfloor n/j \rfloor$. I am trying to prove the identity
$$t(1) + t(2) + \cdots + t(n) = \Big\lfloor \dfrac{n}{1} \Big\rfloor + \Big\lfloor \dfrac{n}{2} \Big\rfloor + \cdots + \Big\lfloor \dfrac{n}{n} \Big\rfloor,$$
where $t(j)$ is the number of divisors of $j$. 
Attempt: The highest power of $k \in \{2, 3, \ldots, n\}$ that divides $n!$ is given by 
$$H_{k} := \Big\lfloor \dfrac{n}{k}\Big\rfloor + \Big\lfloor \dfrac{n}{k^{2}}\Big\rfloor + \cdots + \Big\lfloor\dfrac{n}{k^{N_{c}}}\Big\rfloor,$$
where $N_{c}$ is the maximal integer satisfying $k^{N_{c}} \leq n$. With a little thought, one can see that 
\begin{equation*}
\begin{split}
\Big\lfloor\dfrac{n}{1}\Big\rfloor + \Big\lfloor\dfrac{n}{2}\Big\rfloor + \cdots + \Big\lfloor\dfrac{n}{n}\Big\rfloor &= H_{2} + H_{3} + \cdots + H_{n} + \Big\lfloor\dfrac{n}{1}\Big\rfloor \\
&= H_{2} + H_{3} + \cdots + H_{n} + n.
\end{split}
\end{equation*}
Each $H_{k}$ is a counted in the sum $t(1) + t(2) + \cdots + t(n)$, since each $t(k^{j})$ is a summand for $j \in \{1, 2, \ldots, N_{k}\}$, and also each $t(m)$ is a summand, where $m$ is a multiple of $k$ less than or equal to $n$.
This is where I'm a little stuck: I think that the only other summand in $t(1) + t(2) + \cdots + t(n)$ would be $1 + 1 + \cdots + 1 = n$, since each $t(j)$ counts $1$ as a divisor. My problem is showing that these are the only extra terms in $t(1) + t(2) + \cdots + t(n)$, which would show equality. 
Any hints would be appreciated, or even suggestions of going about it a different way. 
 [/INST] For the  sake of completeness I would  like to point out  that this is
usually done by induction. We seek to show that
$$\sum_{k=1}^n \tau(k)
= \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor.$$
It holds for $n=1:$ $$\tau(1) = \lfloor 1\rfloor.$$
 In the induction step we start from
$$\sum_{k=1}^n \tau(k)
= \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor.$$
to get
$$\tau(n+1) + \sum_{k=1}^n \tau(k)
= \tau(n+1) + \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor.$$
Now $$\tau(n+1) = \sum_{k=1}^{n+1}
\left(\bigg\lfloor \frac{n+1}{k} \bigg\rfloor
- \bigg\lfloor \frac{n}{k} \bigg\rfloor\right)$$
since
$$\bigg\lfloor \frac{n+1}{k} \bigg\rfloor
- \bigg\lfloor \frac{n}{k} \bigg\rfloor
= \begin{cases} 1 \quad\text{if}\quad k|n+1
\\ 0 \quad\text{otherwise}.\end{cases}$$
Therefore we have
$$\sum_{k=1}^{n+1} \tau(k)
= \sum_{k=1}^{n+1}
\left(\bigg\lfloor \frac{n+1}{k} \bigg\rfloor
- \bigg\lfloor \frac{n}{k} \bigg\rfloor\right)
+ \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor
\\ = \sum_{k=1}^{n+1}
\left(\bigg\lfloor \frac{n+1}{k} \bigg\rfloor
- \bigg\lfloor \frac{n}{k} \bigg\rfloor\right)
+ \sum_{k=1}^{n+1} \bigg\lfloor \frac{n}{k} \bigg\rfloor
\\ = \sum_{k=1}^{n+1}
\bigg\lfloor \frac{n+1}{k} \bigg\rfloor$$
and the induction is complete.
</s>