Are These 2 Solutions the Same? 
Suppose the probability that a family has exactly $n$ children is $(1-p)p^n$, $n\ge0$. Assuming each child is equally likely to be a boy or a girl, independently of previous children, find a formula for the probability that a family contains exactly $k$ boys.

My Solution:
Let $X$ be the # of boys in a family, $X\ge0$
Let $Y$ be the # of children in a family, $Y\ge0$
$P(X=k)$
$=\sum_{i=0}^{\infty}P(X=k\,|\,$Y$=i)P($Y$=i)$
$=\sum_{i=0}^{\infty}\frac{i!}{k!(i-k)!}(\frac{1}{2})^k(\frac{1}{2})^{i-k} \cdot (1-p)p^i$
$=(\frac{1-p}{k!})\sum_{0}^{\infty}\frac{i!}{(i-k)!}(\frac{1}{2}p)^i$
Not sure how to simplify it further.
I know my summation of the # of children started with $0$ instead of the "out-of-the-box" thinking of $k$, but those extra cases would just result in a probability of $0$. So in the end, it should be the same as the Textbook Solution:

If so, how to simplify from where I left off to get the final answer in the textbook?
 A: Your answer is incorrect because you can't have more boys than you have children, so all of the values where $k < n$ have probability 0. This is why the books' solution starts at k.
Just for fun, another way to think about the solution is given by solving a recursive equation for $P[k\textrm{ boys}] = f(P[k-1 \textrm{ boys})$ and solving for $k=0$.
Let 
\begin{equation}
P[\textrm{Number of Kids} = n] = (1-p)p^n
\end{equation}
and, assuming the probability of having a boy and girl is equal, 
\begin{equation}
P[k\textrm{ boys}| n \textrm{ children}] = \binom{n}{k} p_b^k(1-p_b)^{n-k} = \binom{n}{k} .5^n
\end{equation}
for $k \leq n$ and $0$ for $k>n$.
Then by the law of total probability: $\sum_{n=0}^\infty P[k\textrm{ boys}| n \textrm{ children}]P[\textrm{Number of Kids} = n]$.
Because $P[k\textrm{ boys}| n \textrm{ children}] =0$ for $k>n$ this reduces to:
\begin{equation}
\sum_{n=k}^\infty P[k\textrm{ boys}| n \textrm{ children}]P[\textrm{Number of Kids} = n] =  (1-p)\sum_{n=k}^\infty \binom{n}{k}(.5p)^n
\end{equation}
Let $q = .5*p$, and let $h(k,q) = \sum_{n=k+1}^\infty \binom{n}{k}q^n$. Then $P[k\textrm{ boys}] = (1-p)[q^k + h(k,q)] $.
Using the identity $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ for $k \leq n-1$:
\begin{equation}
\begin{split}
h(k,q) & = \sum_{n=k+1}^\infty \binom{n}{k}q^n\\
& = \sum_{n=k+1}^\infty [\binom{n-1}{k-1}q^n + \binom{n-1}{k}q^n]\\
& = \sum_{j=k}^\infty [\binom{j}{k-1}q^{j+1} + \binom{j}{k}q^{j+1}]\\
& = q[\sum_{j=k}^\infty \binom{j}{k-1}q^{j} + \sum_{j=k}^\infty \binom{j}{k}q^{j}]\\
& = q[h(k-1,q) + h(k,q) + q^k]\\
\end{split}
\end{equation}
Noting that $\frac{P[k \textrm{ boys}]}{(1-p)} - q^k = h(k,q)$, this gives us that:
\begin{equation}
\begin{split}
h(k,q) & = q[h(k-1,q) + h(k,q) + q^k] \\
\Leftrightarrow \frac{P[k \textrm{ boys}]}{(1-p)} - q^k & = q[\frac{P[k-1 \textrm{ boys}]}{(1-p)} - q^{k-1} + \frac{P[k \textrm{ boys}]}{(1-p)}]\\
\Leftrightarrow P[k \textrm{ boys}]  & = \frac{q}{(1-q)}P[k-1 \textrm{ boys}]\\
\end{split}
\end{equation}
Dividing by $\frac{.5}{.5}$ gives us:
\begin{equation}
P[k \textrm{ boys}]  = \frac{p}{(2-p)}P[k-1 \textrm{ boys}]
\end{equation}
Solving for $k= 0 $ gives us $\sum_{i=0}^\infty (.5p)^i = \frac{1}{1-.5p} = \frac{2}{2-p}$.
Putting the recursion together with $k=0$ gives us:
\begin{equation}
P[k \textrm{ boys}]  = \frac{2p^k}{(2-p)^{k+1}}
\end{equation}
Which is the specified solution.
