Finding the covariance of two random variables knowing the joint probability distribution 
Let $X,Y$ be random variables such that for every $ k \gt 1 $ and for every  $ k \gt m \ge 1$ we get $P[ X = m, Y = k] = \frac {1}{25} (\frac{4}{5})^{k-2}. Cov(X,Y) = ? $

I've written some probabilities to understand better the problem:
$$\begin{array}{c|c|c|c|} 
\text Y/X & \text{1} & \text{2} & \text{3} \\ \hline
\text{2} & \frac{1}{25}  \\ \hline
\text{3} & \frac{4}{125} & \frac{4}{125} \\ \hline
\text{4} & \frac{16}{625} & \frac{16}{625} & \frac{16}{625} \\ \hline
\end{array}$$
In order to find the expected value of Y, we need to compute $ \sum_{k=2}^{\infty} k\cdot P(Y=k) =\frac{1}{25} \sum_{k=2}^{\infty} k\cdot (k-1)\cdot(\frac{4}{5})^{k-2}$.
I dont know how to compute it by myself, so I used wolfram alpha, which said the sum is $\frac {250}{25}$.
https://www.wolframalpha.com/input/?i=%E2%88%91%28k%29%28k-1%29%284%2F5%29%5E%28k-2%29
Looking at the table, we see that, for instance $ P(X=1) = \sum_{k=0}^{\infty} \frac{1}{25}\cdot(\frac{4}{5})^{k}$. So the expected value of $X$ is $ \sum_{k=1}^{\infty}k\cdot P(X=k) = \frac{1}{25}\sum_{k=1}^{\infty}k\cdot \sum_{j=k-1}^{\infty} (\frac{4}{5})^{j}$. I dont have the knowledge to compute this by myself, so I've used wolfram alpha, which says the sum is 5.
Link to wolfram
And to find the expected value of $XY$, $E[XY] = \frac{1}{25}\sum_{k=2}^{\infty}\sum_{m=1}^{k-1}k\cdot m\cdot(\frac{4}{5})^{k-2}$. The result is, according to wolfram, 70.
Link to wolfram
So, in total $Cov(X,Y) = 70 - 5\cdot 10 = 20$
First, it would be nice to know if my answer and way is right.
Second, and more importantly, since I dont know how to compute this sums by my self, and since I haven't been taught to compute such sums, I was thinking there is another way to solve this question. If so, I would be glad to learn.
 A: Your table is good way to visualize the probability masses (even better is to not simplify anything in the table in order to better see the patterns). Your approach is fine, and if you are interested, the kinds of series you entered into wolfram have more general closed forms that are related to the geometric series.
An alternative approach is to observe that $X|Y=y\quad \sim \text{Unif}\{1,...,y-1\},$ so $E[X|Y]=\frac{Y}{2}.$ Thus,
$$\begin{align*}
\text{Cov}(X,Y)&=E[XY]-E[X]E[Y]\\
&=E[E[XY|Y]]-E[E[X|Y]]E[Y] \quad (\text{LIE})\\
&=E[YE[X|Y]]-E[E[X|Y]]E[Y]\\
&=E\left[\frac{Y^2}{2}\right]-E\left[\frac{Y}{2}\right]E[Y]\\
&=\frac{1}{2}\left(E[Y^2]-(E[Y])^2\right)\\
&=\frac{1}{2}\text{Var}(Y).
\end{align*}$$
From your table, the marginal distribution of $Y$ is
$$P(Y=y)=(y-1)\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^{y-2},\quad y\in \{2,3,...\}.$$
Using this, you can work out the variance of $Y$. A shortcut is to note that $Y$ is a shifted negative binomial random variable:
$$Y=Y'+2,\quad Y'\sim NB(r=2,p=4/5).$$
Thus,
$$\text{Cov}(X,Y)=\frac{1}{2}\text{Var}(Y)=\frac{1}{2}\text{Var}(Y'+2)=\frac{1}{2}\text{Var}(Y')=\frac{1}{2}\frac{pr}{(1-p)^2}=\frac{1}{2}\frac{\frac{4}{5}(2)}{\left(1-\frac{4}5{}\right)^2}=20.$$
