Rolling a die - Conditional Probability A die is thrown repeatedly.
Let $X$ ~ First 5 is thrown and $Y$ ~ First 6 is thrown
Calculate $\mathbb{E}(X|Y=3)$
You may use the identity: $\sum_{n=k}^\infty nz^{n-k} = \frac{1}{(1-z)^2}+\frac{k-1}{1-z}$
I know from the definition of expectation, we have:
$\mathbb{E}(X|Y=3) = (1*\frac{1}{5})+(2*\frac{4}{5} * \frac{1}{5}) + (3* \frac{4}{5}* \frac{4}{5} * 0) + (5* \frac{4}{5} * \frac{4}{5} * 1 * \frac{5}{6} * \frac{1}{6}) + (6* \frac{4}{5} * \frac{4}{5} * 1 * \frac{5}{6} * \frac {5}{6} * \frac{1}{6}) + ...$, where every following term, has an extra '$*\frac{5}{6}$' term and constant increases by 1.
However I am unsure of how to apply this to the identity given to find the value of the infinite sum?

While it is not necessary, if the question expects you to use the given identity then here is how you would go about it -
$ \small \displaystyle E(X|Y = 3) = 1 \cdot \frac{1}{5} + 2 \cdot \frac{4 \cdot 1}{5^2} + 4 \cdot \frac{4 \cdot 4}{5^2} \cdot \frac{1}{6} + 5 \cdot \frac{4 \cdot 4}{5^2} \cdot \frac{5 \cdot 1}{6^2} + ...$
$ \small \displaystyle  = \frac{13}{25} + \frac{4 \cdot 4}{5^2} \cdot \frac{1}{6} \left (4 \left(\frac{5}{6}\right)^0 + 5 \cdot \left(\frac{5}{6}\right)^1 + ...\right)$
Now note that comparing the below against the given identity,
$ \small \displaystyle 4 \left(\frac{5}{6}\right)^0 + 5 \cdot \left(\frac{5}{6}\right)^1 + ...$
$ \displaystyle \small z = \frac{5}{6}, k = 4$ and $\sum \limits_{n=k}^\infty nz^{n-k} = 54$
and we get, $ \displaystyle \small E(X|Y = 3) = \frac{157}{25}$
