How do I create a bash script that sums any number of command line arguments?

No need for bash, plain sh will do as well:

#! /bin/sh - 
IFS=+; echo "$(($*))"

$* in POSIX shells, expands to the list of positional parameters (in this case, the arguments to the script) separated by the first character of $IFS (or space if $IFS is unset or nothing if $IFS is empty). $((...)) is the shell internal arithmetic expansion operator (note that it supports decimal, octal and hexadecimal numbers)

If you need floating point support, that's where you'll need a different shell like ksh93 or zsh (not bash as bash only supports integer arithmetic), though you could also use awk:

#! /usr/bin/awk -f
BEGIN {t=0; for (i in ARGV) t+=ARGV[i]; print t}

That will use double type numbers as implemented by your system. The input numbers must be decimal floating point or engineering notation in the English style (floating point delimiter is the period character regardless of the locale).

A non-looping variant:

{ printf %d+ "$@"; echo 0; } | bc

Example

Put the above in a script file, sum.

#!/bin/bash

{ printf %d+ "$@"; echo 0; } | bc

Run it like so:

$ ./sum 4
4
$ ./sum 4 4 5
13

You can use the following bash function:

sum() {
    local sum=0
    for arg in "$@"; do
        (( sum += arg ))
    done   
    echo $sum
}