$\sum_{k = 1}^{\infty} k\lfloor\frac{n}{k} \rfloor = 1 + \sum_{k = 1}^n \sigma_1(n)$

Question

For any $f: \Bbb{N} \to \Bbb{Z}$ there exists a unique transformed function $F:\Bbb{N} \to \Bbb{Z}$ such that:

$$ f(n) = \sum_{k = 1}^{\infty}F_k\lfloor\frac{n}{k}\rfloor $$

For example, set $F_1 = f(1)$. Then at $n = 2$ we want to have $F_1\cdot n + F_2\lfloor\frac{n}{1}\rfloor = f(2)$, so we let $F_2 := f(2) - \sum_{k=1}^2 \lfloor\frac{2}{k} \rfloor$, similarly set $F_n := f(n) - \sum_{k=1}^{n-1} \lfloor\frac{n}{k} \rfloor$. Clearly the coefficeints $F_n$ are uniquely determined.

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Knowing that, how can we prove that, if $T(f) := F$ in the above, $T(\text{A160664}(n))(n) = \text{id}$?

https://oeis.org/A160664

That is:

$$ \sum_{k = 1}^{\infty} k\lfloor\frac{n}{k} \rfloor = 1 + \sum_{k = 1}^n \sigma_1(n) $$

Answer 1

Actually, the $1$ on the right is spurious, as you can easily verify from $n=1$ (LHS = 1) or $n=2$ (LHS = 4). Unless you somehow have defined $\sigma_1(1) := 0$?

The terms on the left are identically zero when $k > n$, so the LHS is equal to $$\sum_{k=1}^n k \lfloor \frac{n}{k} \rfloor = \sum_{k=1}^n k \sum_{d=1}^n \mathbf 1_{dk\le n} = \sum_{k=1}^n k \sum_{d=1}^n \sum_{r=1}^n \mathbf 1_{dk=r},$$ where $\mathbf 1_{P}$ is the indicator function that takes the value $1$ if the proposition $P$ is true and $0$ otherwise.

Now switch the order of the sums to get $$\sum_{r=1}^n \sum_{k=1}^n k \sum_{d=1}^n \mathbf 1_{dk = r} = \sum_{r=1}^n \sum_{k=1}^n k \cdot \mathbf 1_{k \mid r} = \sum_{r=1}^n \sum_{k : k \mid r} k = \sum_{r=1}^n \sigma_1(r).$$