[proofplan] We realize the Brownian bridge as [Brownian motion](/page/Brownian%20Motion) conditioned to return to zero at time $1$. The event that the bridge remains in $(-a,a)$ is then computed by the Brownian transition density killed at the endpoints $\pm a$, divided by the unrestricted transition density from $0$ to $0$. The reflection principle gives an explicit image-series formula for the killed density. After dividing by the unrestricted density and separating even and odd image terms, the stated alternating series follows. Finally, continuity of the supremum distribution lets us replace the open event by the closed event with $\le a$. [/proofplan] [step:Reduce the bridge event to a killed Brownian transition density] Let $(W_t)_{0\le t\le 1}$ be standard Brownian motion started at $0$, with transition density \begin{align*} p_t(x,y)=\frac{1}{\sqrt{2\pi t}}\exp\left(-\frac{(y-x)^2}{2t}\right). \end{align*} Let $I=(-a,a)$ and let $p_{t,I}(x,y)$ denote the transition density of Brownian motion killed on exiting $I$. A standard construction of the Brownian bridge says that the law of $B$ is the law of $W$ conditioned on $W_1=0$. Therefore \begin{align*} \mathbb P\left(\sup_{0\le t\le 1}|B(t)|<a\right) =\frac{p_{1,I}(0,0)}{p_1(0,0)}. \end{align*} Indeed, conditioning on the endpoint $W_1=0$, the numerator is the joint density contribution from paths that stay inside $I$ up to time $1$ and end at $0$, while the denominator is the unrestricted endpoint density. [/step] [step:Use the reflection principle to compute the killed density] For Brownian motion killed on an interval $(0,L)$, write $p_{t,(0,L)}(x,y)$ for its killed transition density. The reflection principle gives the image formula \begin{align*} p_{t,(0,L)}(x,y) =\sum_{m\in\mathbb Z}\left(p_t(x,y+2mL)-p_t(x,-y+2mL)\right). \end{align*} Translate $I=(-a,a)$ to $(0,2a)$ by adding $a$ to space, and take $x=y=a$ in the translated interval. With $L=2a$ and $t=1$, this gives \begin{align*} p_{1,I}(0,0) =\frac{1}{\sqrt{2\pi}}\sum_{m\in\mathbb Z} \left(e^{-8m^2a^2}-e^{-2(2m+1)^2a^2}\right). \end{align*} Since $p_1(0,0)=1/\sqrt{2\pi}$, the ratio from the previous step is \begin{align*} \sum_{m\in\mathbb Z} \left(e^{-8m^2a^2}-e^{-2(2m+1)^2a^2}\right). \end{align*} [/step] [step:Rewrite the image sum as the Kolmogorov alternating series] Separate the first image sum into the term $m=0$ and the symmetric pairs $m$ and $-m$: \begin{align*} \sum_{m\in\mathbb Z}e^{-8m^2a^2} =1+2\sum_{m=1}^{\infty}e^{-8m^2a^2}. \end{align*} The second image sum has the same value for $m=j$ and $m=-j-1$, so \begin{align*} \sum_{m\in\mathbb Z}e^{-2(2m+1)^2a^2} =2\sum_{j=0}^{\infty}e^{-2(2j+1)^2a^2}. \end{align*} Thus \begin{align*} \mathbb P\left(\sup_{0\le t\le 1}|B(t)|<a\right) =1+2\sum_{m=1}^{\infty}e^{-2(2m)^2a^2} -2\sum_{j=0}^{\infty}e^{-2(2j+1)^2a^2}. \end{align*} Combining the even positive integers and odd positive integers into one alternating sum gives \begin{align*} \mathbb P\left(\sup_{0\le t\le 1}|B(t)|<a\right) =1+2\sum_{k=1}^{\infty}(-1)^k e^{-2k^2a^2} =1-2\sum_{k=1}^{\infty}(-1)^{k-1}e^{-2k^2a^2}. \end{align*} [/step] [step:Pass from the open boundary event to the closed boundary event] The Brownian bridge has continuous sample paths, so \begin{align*} X:=\sup_{0\le t\le 1}|B(t)| \end{align*} is a real-valued [random variable](/page/Random%20Variable). The formula just proved for the open event is continuous as a function of the boundary level, because the series converges uniformly on compact subintervals of $(0,\infty)$ by comparison with $\sum_{k\ge1}e^{-c k^2}$. For $c>a$, the events $\{X<c\}$ decrease as $c\downarrow a$, and \begin{align*} \bigcap_{c>a}\{X<c\}=\{X\le a\}. \end{align*} Continuity from above for probabilities gives \begin{align*} \mathbb P(X\le a)=\lim_{c\downarrow a}\mathbb P(X<c). \end{align*} Applying the open-boundary formula at level $c$ and then taking $c\downarrow a$ gives \begin{align*} \mathbb P\left(\sup_{0\le t\le 1}|B(t)|\le a\right) =1-2\sum_{k=1}^{\infty}(-1)^{k-1}e^{-2k^2a^2}. \end{align*} [guided] The killed-density computation gives the probability of the strict event. Put \begin{align*} X=\sup_{0\le t\le 1}|B(t)|. \end{align*} For every $c>a$, the open-boundary formula gives \begin{align*} \mathbb P(X<c)=1-2\sum_{k=1}^{\infty}(-1)^{k-1}e^{-2k^2c^2}. \end{align*} As $c$ decreases to $a$, the events $\{X<c\}$ decrease, and their intersection is \begin{align*} \bigcap_{c>a}\{X<c\}=\{X\le a\}. \end{align*} Therefore continuity from above gives \begin{align*} \mathbb P(X\le a)=\lim_{c\downarrow a}\mathbb P(X<c). \end{align*} The series on the right is continuous in $c>0$ by [uniform convergence](/page/Uniform%20Convergence) on compact subintervals of $(0,\infty)$, so taking $c\downarrow a$ gives \begin{align*} \mathbb P(X\le a)=1-2\sum_{k=1}^{\infty}(-1)^{k-1}e^{-2k^2a^2}. \end{align*} [/guided] [/step]