a) $1/999 = 0.001001001\ldots = 0.\overline{001}$
Either of these is fairly standard notation. The overline
format $0.\overline{001}$ is a little more explicit,
so I think it would be preferred.
b) $.001 + .000001 + .000000001 + \cdots$
This denotes an infinite sequence in the way that
$1, 2, 3, \ldots$ indicates an infinite sequence:
a little informally, but I think it is recognizable as a
mathematical notation.
The main problem with this one is it is hard to count the zeros.
One way to help see how big each number is is to consider a power of
$0.001$, like this:
$$0.001^1 + 0.001^2 + 0.001^3 + \cdots.$$
Using the $\sum$ summation notation, this is
$$\sum_{k=1}^\infty 0.001^k.$$
Alternatively, you could write this as
$$10^{-3} + 10^{-6} + 10^{-9} + \cdots$$
but now it is starting to look like
c) $1 \cdot 10^{-(3 \cdot (n+0))} + 1 \cdot 10^{-(3 \cdot (n+1))}
+ 1 \cdot 10^{-(3 \cdot (n+2))} + \cdots$
But there's a problem here: what is $n$? If you set $n=1$ then this evaluates
to $10^{-3 \cdot 1} + 10^{-3 \cdot 2} + 10^{-3 \cdot 3} + \cdots$
which is OK. If you set $n$ to any other value then the sum will
not add up to $1/999$.
In $\sum$ notation, this series is
$$\sum_{k=0}^{\infty} 10^{-3 \cdot (n + k)}$$
and we can rewrite it as follows:
$$\sum_{k=0}^{\infty} 10^{-3 \cdot (n + k)}
= 10^{-3n} \sum_{k=0}^{\infty} \left( 10^{-3} \right)^k
= 10^{-3n} \left( 1 + 0.001 + 0.000001 + \cdots \right).
$$
d) $\displaystyle\sum_{k=1}^{\infty} \frac{(10^{\lfloor\strut\log_{10}(p)\rfloor+1} - p)^{k-1}}{(10^{\lfloor\strut\log_{10}(p)\rfloor+1})^k}$
This looks really complicated, but
let's try setting $N = 10^{\lfloor\strut\log_{10}(p)\rfloor+1}$.
Then
\begin{align}
\sum_{k=1}^{\infty} \frac{(10^{\lfloor\strut\log_{10}(p)\rfloor+1} - p)^{k-1}}{(10^{\lfloor\strut\log_{10}(p)\rfloor+1})^k}
& = \sum_{k=1}^{\infty} \frac{(N - p)^{k-1}}{N^k} \\
& = \sum_{k=0}^{\infty} \frac{(N - p)^k}{N^{k+1}} \\
& = \frac 1N \sum_{k=0}^{\infty} \left(1 - \frac pN\right)^k \\
\end{align}
The sum in this last equation is a geometric series $\sum r^k$
with ratio $r = \left(1 - \frac pN\right)$.
The limit of such a series starting from the $k=0$ term
is $\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}$.
Applying this result, we have
$$\sum_{k=0}^{\infty} \left(1 - \frac pN\right)^k
= \frac{1}{1 - \left(1 - \frac pN\right)}$$
and therefore
\begin{align}
\sum_{k=1}^{\infty} \frac{(10^{\lfloor\strut\log_{10}(p)\rfloor+1} - p)^{k-1}}{(10^{\lfloor\strut\log_{10}(p)\rfloor+1})^k}
& = \frac 1N \left(\frac{1}{1 - \left(1 - \frac pN\right)}\right) \\
& = \frac 1N \left(\frac Np\right) \\
& = \frac 1p.
\end{align}
So it turns out it doesn't make much difference what $N$ is.
Edit: the rest of this answer has been revised to reflect upon
the related question, Doubling sequences of the cyclic decimal parts of the fraction numbers.
Now, it turns out that in the case where $p = 999$,
the series in part d) is term-for-term the same as the series in part b),
because $10^{\lfloor\strut\log_{10}(999)\rfloor+1} - 999 = 1$.
The sum then comes down to
$$\sum_{k=1}^{\infty}
\frac{1^{k-1}}{(10^{\lfloor\strut\log_{10}(999)\rfloor+1})^k}
= \sum_{k=1}^{\infty} \frac{1}{1000^k}
= \sum_{k=1}^{\infty} 0.001^k$$
which is a more explicit way of writing the series
$0.001^1 + 0.001^2 + 0.001^3 + \cdots$.
But as the calculations above showed,
the formula for part d) works even when $p$ is not all $9$s
(that is, even when $p \neq 16^n - 1$ for any integer $n$).
For example, take $p = 98$.
The formula then says
$$\frac{1}{98}
= \sum_{k=1}^{\infty} \frac{(100 - 98)^{k - 1}}{100^k}
= \sum_{k=1}^{\infty} \frac{2^{k-1}}{100^k}
= 0.01 + 0.0002 + 0.000004 + 0.00000008 + \cdots$$
which well illustrates the "powers of $2$" pattern within the
repeating $42$-digit sequence in
$$\frac{1}{98} = 0.0\overline{102040816326530612244897959183673469387755}.$$
More could be said along these lines, but the question
Doubling sequences of the cyclic decimal parts of the fraction numbers
is a better context for that.
0.\overline{001}in the (a) part? – Asaf Karagila Jul 23 '15 at 15:00