Gulp task not finishing - gulp

Question

Being new to gulp. I have the follwing task. I need to spawn a webserver and then another script has to run some stuff against this webserver. I am currently struggling with finishing the script because the browser-sync task does not finish and prevents the script from exit.
'use strict';
const browserSync = require('browser-sync').create();
const cp = require('child_process');
const minimalcss = require('minimalcss');
const gulp = require('gulp');
const clean = require('gulp-clean');
const sourceDir = "_site/";
const deployDir = "public/";
// build the mkdocs Site
gulp.task('build', function() {
return cp.exec('pipenv run mkdocs build --site-dir ' + sourceDir);
});
// Delete _deploy directory first
gulp.task('prepare', function() {
return gulp.src(deployDir, {read: false, allowEmpty: true})
.pipe(clean());
});
// Delete _deploy directory again // just for testing
gulp.task('cleanup', function() {
return gulp.src(deployDir, {read: false, allowEmpty: true})
.pipe(clean());
});
// does not lead to anything, just for testing
gulp.task('inlinecriticalCSS', function(done) {
minimalcss
.minimize({ urls: ['http://localhost:9999/' + 'index.html'] })
.then(result => {
console.log('OUTPUT', result.finalCss.length, result.finalCss);
})
.catch(error => {
console.error(`Failed the minimize CSS: ${error}`);
});
done();
});
// webserver
gulp.task('serve', (done) => {
browserSync.init({
port: 9999,
server: {
baseDir: sourceDir
}
});
done();
});
// default sequence
gulp.task('default', gulp.series(
'prepare',
'build',
'serve',
'inlinecriticalCSS',
'cleanup')
);