Any technique to show that $P\left\{X\geq 2\lambda\right\}\leq (e/4)^{\lambda}$

Question

Let $X$ be a random variable having the poisson distribution with parameter $\lambda$. Show that $$P\left\{X\geq 2\lambda\right\}\leq (e/4)^{\lambda}$$I have been stuck in this problem, and I can't seem to get the trick in proving this. Well before this problem, I was able to show that $P\left\{X\geq 2\lambda\right\}\leq 1/\lambda$ using Chebychev's inequality. But in this problem, I really don't know what to do.I tried to bound $1/\lambda$ with $(e/4)^\lambda$ but this is not always true. Any suggestions?

Answer 1

Hint:

$$P(X \geq 2 \lambda)=e^{-\lambda} \sum_{k=N}^\infty \frac{\lambda^k}{k!} = e^{-\lambda} \left ( e^\lambda - \sum_{k=0}^{N-1} \frac{\lambda^k}{k!} \right )$$

where $N=\lceil 2\lambda \rceil$. The sum is a finite truncation of the Maclaurin series which converges to $e^\lambda$. Do you know how to estimate the truncation error for such a series?