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. 
 A: It really helps to write $t(1)+t(2)+\dotsb+t(n)$ in a triangular array: 

1 + 
1 + 1 +
1 + 0 + 1 +
1 + 1 + 0 + 1 +
1 + 0 + 0 + 0 + 1 +
1 + 1 + 1 + 0 + 0 + 1 +
1 + 0 + 0 + 0 + 0 + 0 + 1 +
1 + 1 + 0 + 1 + 0 + 0 + 0 + 1 + 
1 + 0 + 1 + 0 + 0 + 0 + 0 + 0 + 1 + ...

Row $k$ is $t(k)$ written in a way that indicates which numbers from 1 to $k$ divide $k$. What do you see vertically?
As a final hint, note that $\lfloor n/d \rfloor$ counts the number of multiples of $d$ not exceeding $n$.

Since you have resolved your issue, let me explain the above. 
We have
$$\sum_{k=1}^n t(k) = \sum_{k=1}^n \sum_{\substack{1 \le d \le n \\ d|k}} 1$$
and we want to swap the order of summation. Well, as $k$ and $d$ range over $1$ to $n$, you add $1$ every time $d$ divides $k$. This is the same as adding $1$ every time $k$ is a multiple of $d$. For a fixed $d$, how many multiples of $d$ are there between $1$ and $n$? Precisely $\lfloor n/d \rfloor$ (they are $d$, $2d$, $3d$, ..., $md \le n < (m+1)d$). 
So $$\sum_{k=1}^n \sum_{\substack{1 \le d \le n \\ d|k}} 1 = \sum_{d=1}^n \sum_{\substack{1 \le k \le n\\ k = md}} 1 = \sum_{d=1}^n \Big\lfloor \frac{n}{d} \Big\rfloor$$
and the proof is complete. 
A: 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.
A: it is the basic theory of Dirichlet convolution  :
$$\zeta(s) = \sum_{n=1}^\infty n^{-s}$$ $$\zeta(s)^2 = (\sum_{n=1}^\infty n^{-s})(\sum_{m=1}^\infty m^{-s} )= \sum_{n=1}^\infty \sum_{m=1}^\infty n^{-s} m^{-s}= \sum_{n=1}^\infty n^{-s} \sum_{d|n} 1 = \sum_{n=1}^\infty \tau(n) n^{-s}$$ there is talso the Perron formula : 
$$\sum_{n=1}^\infty a_n n^{-s} = \int_{1-\epsilon}^\infty (\sum_{n=1}^\infty a_n \delta(x-n)) x^{-s} dx = s \int_1^\infty (\sum_{n \le x} a_n) x^{-s-1}dx$$ (integration by part)
hence 
$$\zeta(s) =  s \int_1^\infty (\sum_{n \le x}1) x^{-s-1} dx = s \int_1^\infty  \lfloor x \rfloor x^{-s-1} dx$$
$$\zeta(s)^2 =  s \int_1^\infty (\sum_{n \le x}\tau(n)) x^{-s-1} dx$$
but you can also write 
$$\zeta(s)^2 = (\sum_{n=1}^\infty n^{-s} )s \int_1^\infty  \lfloor x \rfloor x^{-s-1} dx = s \int_1^\infty (\sum_{n=1}^\infty \lfloor x/n \rfloor) x^{-s-1}dx$$
(since $\int_0^\infty f(x/n) x^{-s-1}dx = n^{-s} \int_0^\infty f(y) y^{-s-1}dy$ with $x= ny$)
i.e.
$$\sum_{n \le x}\tau(n) = \sum_{n=1}^\infty \lfloor x/n \rfloor  = \sum_{n \le x} \lfloor x/n \rfloor$$
