Distribution of $\max_{t \in [0,1]} |W_t|$ for Brownian motion 
For a standard Brownian motion $\{W_t, t\geq 0\}$, find $\mathbb{P}(\max_{ t \in [0,1]}|W_t| <x)$.

Page 79-80 of Billingsley, P., Convergence of probability measures, New York-London-Sydney-Toronto: John Wiley and Sons, Inc. XII, 253 p. (1968). ZBL0172.21201. says:
$\mathbb{P}(\max_{ t \in [0,1]}|W_t| <x)=1-\frac{4}{\pi}\sum \frac{(-1)^{k}}{2k+1} \exp\left(-\frac{\pi^2 (2k+1)^2}{8 x^2}\right)$
I think it is not correct. I plotted the series $k=100$ and  $x\in[0,10]$. It is really weird. The probability is always larger than $1$ and it goes to $1.2$ !!! Can you help me find the problem?


The formula should be 
$$P(\max_{t\in [0,1]} |w_t| < x)=\frac{4}{π} \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \exp\left\{-\frac{π^2(2n+1)^2}{8x^2}\right\}$$
Here is a plot of the resulting function:

