how to read file from line x to the end of a file in bash

tail -n +2 file.csv

From the man page:

-n, --lines=N
     output the last N lines, instead of the last 10
...

If the first character of N (the number of bytes or lines)  is  a  '+',
print  beginning with the Nth item from the start of each file, other-
wise, print the last N items in the file.

In English this means that:

tail -n 100 prints the last 100 lines

tail -n +100 prints all lines starting from line 100

Simple solution with sed:

sed -n '2,$p' <thefile

where 2 is the number of line you wish to read from.

Or else (pure bash)...

{ for ((i=1;i--;));do read;done;while read line;do echo $line;done } < file.csv

Better written:

linesToSkip=1
{
    for ((i=$linesToSkip;i--;)) ;do
        read
        done
    while read line ;do
        echo $line
        done
} < file.csv

This work even if linesToSkip == 0 or linesToSkip > file.csv's number of lines

Edit:

Changed () for {} as gniourf_gniourf enjoin me to consider: First syntax generate a sub-shell, whille {} don't.

of course, for skipping only one line (as original question's title), the loop for (i=1;i--;));do read;done could be simply replaced by read:

{ read;while read line;do echo $line;done } < file.csv

There are many solutions to this. One of my favorite is:

(head -2 > /dev/null; whatever_you_want_to_do) < file.txt

You can also use tail to skip the lines you want:

tail -n +2 file.txt | whatever_you_want_to_do