Define
$$f_k=\sum_{n=1}^{k}\delta_n$$
Show that the limit of $f_k$ exists as a distribution. Find a locally integrable function $F$ such that the distributional derivative of $F$ is the limit of $f_k$
This is what I've managed so far:
$\delta_n$ is the distributional derivative of $H(x-n)$
let $F=\sum_{n=1}^{\infty}H(x-n)= \begin{cases} \lfloor x \rfloor & \text{if $x\geq0$} \\ 0 & \text{otherwise} \end{cases}$
$-\langle F, \phi'\rangle=-\int_{\mathbb{R}} (\sum_{n=1}^{\infty}H(x-n))\phi'(x) \ dx = -\sum_{n=1}^{\infty} (\int_{\mathbb{R}}H(x-n)\phi'(x) \ dx)$
Integrate by parts:
$-\sum_{n=1}^{\infty} (\int_{\mathbb{R}}H(x-n)\phi'(x) \ dx) = -\sum_{n=1}^{\infty}(-\int_{\mathbb{R}}\delta_n\phi(x) \ dx \ + \ [H(x-n)\phi(x)]_{x=-\infty}^{x=\infty})$
The boundary terms go to $0$ because of compact support
$\sum_{n=1}^{\infty}(\int_{\mathbb{R}}\delta_n\phi(x) \ dx) = \int_{\mathbb{R}}(\sum_{n=1}^{\infty}\delta_n)\phi(x) \ dx=\langle f_k,\phi \rangle$
Any help is appreciated!