I'll answer how they derived the mean or expected value of the distribution, and then I'll leave it to you to derive the variance (it'll involve a similar sort of method).
This is assuming $i=0$ (you can easily prove that for an arbitrary $i$ it just amounts to adding $i$ at the end). Think of this as just 'shifting' the distribution $i$ units to the right (or left).
For a random variable $X$ we define the 'mean' or expected value of $X$ as
$\sum_{\forall x} P(X=x)x$ where $P(X=x)$ is the probability that the random variable $X$ takes on the value $x$.
Lets first define the random variable in this scenario, here it is:
$X = {\frac{i}{2}}$, $\forall i \in +/- \{1,2,3,...N\}$
We can also define a probability function for $X$ as
$P(X=x) = {\frac{p}{N}}\\ \forall i \in + \{1,2,...,N\}$
$P(X=x) = {\frac{1-p}{N}} \\ \forall i\in - \{1,2,...,N\}$
So
$E(x) = \sum_{\forall x}P(X=x)x = \sum_{i \in \{1,2,...,n\}}\frac{p}{N}\frac{i}{2} + \sum_{i \in \{-1,-2,...,-n\}}\frac{1-p}{N}\frac{i}{2} $
Also,
$\sum_{i \in \{1,2,...,n\}}\frac{p}{N}\frac{i}{2} = \frac{1}{2}\frac{p}{N} \sum_{i \in \{1,2,...,n\}}{i}$
But $ \sum_{i \in \{1,2,...,n\}}{i}$ is just the sum of the first $N$ natural numbers, which is known to be $\LARGE \frac{N(N+1)}{2}$
So $\sum_{i \in \{1,2,...,n\}}\frac{p}{N}\frac{i}{2} = \frac{1}{2}\frac{p}{N} \sum_{i \in \{1,2,...,n\}}{i} = \frac{1}{2}\frac{p}{N}\frac{N(N+1)}{2} = \frac{p(N+1)}{4}$
Similarly, we can also show that $\sum_{i \in \{-1,-2,...,-n\}}\frac{1-p}{N}\frac{i}{2} = \frac{p-1(N+1)}{4}$
And so $E(x) = \sum_{\forall x}P(X=x)x = \sum_{i \in \{1,2,...,n\}}\frac{p}{N}\frac{i}{2} + \sum_{i \in \{-1,-2,...,-n\}}\frac{1-p}{N}\frac{i}{2} = \frac{p(N+1)}{4} + \frac{(p-1)(N+1)}{4} = \frac{2p-1(N+1)}{4}$
Already you have some confusion on the definition of expected value. You write
$$"E(X)=cp+cp^2/2+\dotsc+cp^k/k"$$
First, was it a typo to leave off the rest of the infinite sum? Second, you aren't even abiding by the definition of expectation. We know $P(X=k)=cp^k/k$ for $k=1,2,\dotsc$. So, by definition,
$$E(X)=\sum_{k=1}^\infty kP(X=k)=c\sum_{k=1}^\infty k \cdot \frac{p^k}{k}=c\sum_{k=1}^\infty p^k$$
$$=c(p+p^2+p^3+\dotsc)=cp(1+p+p^2+\dotsc),$$
and if you recognize the final series and are aware of its closed form expression then you can easily conclude. If not, then you need to review geometric series.
The other answer shows you how to find $c$ so I have omitted that part. Finding the variance amounts to computing
$$E(X^2)=c \sum_{k=1}^\infty kp^k$$ and using the fact that the variance is equal to $E(X^2)-E(X)^2$. To compute the new series, pull out a $p$ and recognize it as the derivative of the series $p+p^2+\cdots$. Is this doable?
Comment if you need clarification.
Best Answer
Apologies for the confusion, I've found the answer thanks to a hint from Fede Poncio in the comment on the original question. Also, the second part of my question in the Edit section originates from my initial lack of understanding of geometric series.
\begin{align} E(X) & = \sum_{k=0}^\infty k \times c \frac{p^k}{k}\\ & = c\sum_{k=1}^\infty p^k \\ & = c\sum_{k=1}^\infty p \times p^{k-1} \\ & = -\frac{p}{\log(1-p)}\sum_{k=1}^\infty p^{k-1} \\ & = -\frac{p}{\log(1-p)}\frac{1}{1-p} \\ & = -\frac{p}{(1-p)\log(1-p)} \end{align}
I forgot that geometric series should start from the case of $x^0$, not $x^1$. And since I changed the base of the summation from $k=0$ to $k=1$, the appropriate changes to the ratio $p$ were also made.
\begin{align} E(X^2) & = \sum_{k=0}^\infty k^2 \times c\frac{p^k}{k} \\ & = cp\sum_{k=1}^\infty kp^{k-1} \\ & = -\frac{1}{\log(1-p)} \times \frac{p}{(1-p)^2} \\ & = -\frac{p}{(1-p)^2\log(1-p)} \\ \end{align}
\begin{align} Var(X) & = E(X^2)\ -\ (E(X))^2 \\ & = -\frac{p}{(1-p)^2\log(1-p)}\ -\ \frac{p^2}{(1-p)^2(\log(1-p))^2} \\ & = -\frac{p\log(1-p)\ +\ p^2}{(1-p)^2(\log(1-p))^2} \\ & = -\frac{p^2\ +\ p\log(1-p)}{(1-p)^2(\log(1-p))^2} \end{align}
You can verify that these are correct from the Wikipedia page for the Logarithmic distribution.
For anybody who's also wondering how I initially did about the series:
$$\sum_{k=1}^\infty kp^{k-1}$$
as Fede Poncio pointed out, it helps if you observe that this is the first derivative of the series:
\begin{align} &\sum_{k=0}^\infty p^k = \frac{1}{1-p} \\ &\sum_{k=1}^\infty kp^{k-1} = \frac{1}{(1-p)^2} \\ \end{align}