Telling which sum is greater, the odd or even powers of a Poisson random variable Let $X$ be a Poisson random variable with parameter $\lambda$, then we have that 
$$P(X=k)=e^{-\lambda} \frac{\lambda^k}{k!}$$
and we know that $\sum_{k=0}^{\infty}P(X=k)=1$, now, this can be splitted into 
$$\sum_{k=0}^{\infty}P(X=k)=\sum_{k=0}^\infty P(X=2k) + \sum_{k=0}^\infty P(X=2k+1)=1$$
then we compute 
$$\sum_{k=0}^\infty P(X=2k)= e^{-\lambda}\sum_{k=0}^\infty \frac{\lambda^{2i}}{(2i)!} = e^{-\lambda} \cosh(\lambda)$$
therefore 
$$\sum_{k=0}^\infty P(X=2k+1)=1-e^{-\lambda} \cosh(\lambda)=\frac{1}{2}(1-e^{-2\lambda})$$
from this we conclude that the even sum is greater that the odd sum.
My question is, Are there more approaches to do this? only using the following 
$$P(X=k+1)=\frac{\lambda}{k+1}P(X=k)$$
I was thinking to use a Gauss's trick here but I can't figure out how.
Thanks a lot in advance

Using your methods, we can find the odd sum in the same way as we find the even sum:
$$\mathbb P(X \text{ odd}) = \mathrm e^{-\lambda} \sum_{k=1}^\infty \frac{\lambda^k}{k!}=\mathrm e^{-\lambda} \sinh \lambda$$
And this shows quite easily that the even probability is greater.
From an intuitive point of view, one may ask why an even result should be more likely than an odd one. This essentially stems from the smallest possible value that $X$ may take being 0. As $\lambda$ increases, we find that the difference between the probabilities decreases exponentially and that the bias between odd and even results only manifests itself when the expectation of $X$ is close to 0.
Ok. Here is how to do it under some conditions with the relation you gave at the end of your question. First we define:
$$p_k = \mathbf P(X=k).$$
Thus, $$p_{k+1} = \frac\lambda{k+1} p_{k}.$$ Now let $$s_k =p_k+p_{k+2}+p_{k+4}+\cdots$$
Combining the two, we see that
$$s_{k+1} = \lambda((k+1)^{-1}p_k +(k+3)^{-1}p_{k+2}+\cdots)$$
But as $k\ge0,$ we have:
$$s_{k+1}\le\frac\lambda{k+1}(p_k+p_{k+2}+\cdots).$$
Thus $$s_1\le\lambda s_0$$
This gives the desired result for $\lambda>1$. I'll leave it up to you to see if you can finish it off
