Poisson conditional expectation Random variable X I can't seem to work this out.

Let $X$ be a Poisson random variable with parameter $\lambda$. Find the conditional expected value of $X$ given $X$ is odd.

 A: EDIT: Sorry about the error, got it corrected now. Thanks, Dilip and Did.
What you need is conditional expectation $\mathbf{E}[X|F]=\frac{E[X \cap F]}{P(F)}$, where $F$ is the set of all odd integers:
$$
\mathbf{E}[X \cap F]= \sum_{k=1}^{\infty}\mathbf{E}[X \cap F_k]P[F_k]=1 \cdot e^{-\lambda} \lambda + 3 \cdot \frac{e^{-\lambda}{\lambda^3}}{3!} + \ldots \\
= \sum_{k=0}^{\infty} (2k+1) \frac{e^{-\lambda} \lambda^{2k+1}}{(2k+1)!}=e^{-\lambda} \lambda \sum_{k=1}^{\infty}\frac{\lambda^{2k}}{(2k)!}=e^{-\lambda}\lambda \cosh \lambda
$$
At the same time. $P(F)=e^{-\lambda} \sinh \lambda$. Hence, $\mathbf{E}[X|F]=\frac{\mathbf{E}[X \cap F]}{P(F)} = \lambda \coth \lambda$
A: The probability of the event $A = \{X~\mathrm{is~odd}\}$ is
$$\begin{align}P(A) &= P\{X=1\} + P\{X=3\} + P\{X=5\} + \cdots\\
&= \lambda\exp(-\lambda) + \frac{\lambda^3\exp(-\lambda)}{3!} 
+ \frac{\lambda^5\exp(-\lambda)}{5!} + \cdots\\
&= \exp(-\lambda)\left[\lambda + \frac{\lambda^3}{3!} 
+ \frac{\lambda^5}{5!} + \cdots\right]\\
&= \exp(-\lambda)\sinh(\lambda).
\end{align}$$
Thus, the conditional probability mass function of $X$ given the
event $A$ that $X$ is odd is
$$\begin{align}
p_{X\mid A}(n) &= P\{X=n\mid A\} = \frac{P\left(\{X=n\}\cap A\right)}{P(A)}\\
&= \begin{cases}\frac{P(X=n\}}{P(A)}, & n ~ \text{odd}\\
0, & n~ \text{even}\end{cases}\\
&= \begin{cases}\frac{\lambda^n}{n!\sinh(\lambda)}, & n ~ \text{odd}\\
0, & n~ \text{even}\end{cases}
\end{align}$$
The expected value of $X$ given $A$ is thus
$$\begin{align}
E[X\mid A] &= 1\cdot \frac{\lambda}{1!\sinh(\lambda)}
+ 3\cdot \frac{\lambda^3}{3!\sinh(\lambda)} + 5\cdot \frac{\lambda^5}{5!\sinh(\lambda)}
+ \cdots \\
&= \frac{1}{\sinh(\lambda)}\left[1\cdot \frac{\lambda}{1!}
+ 3\cdot \frac{\lambda^3}{3!} + 5\cdot \frac{\lambda^5}{5!}
+ \cdots\right].
\end{align}$$
I will leave the summation of the series to you. See, for example,
Alex's answer for ideas, but be careful in applying his approach;
I think there are some typographical errors in it and the correct answer
is $E[X\mid A] = \lambda\coth(\lambda)$.
Edit Alex has corrected his calculations and now gets the same
value for $E[X\mid A]$ as I gave above.
