DOC: update the Period.dayofweek attribute docstring
Checklist for the pandas documentation sprint (ignore this if you are doing
an unrelated PR):

- [x] PR title is "DOC: update the <your-function-or-method> docstring"
- [x] The validation script passes: `scripts/validate_docstrings.py <your-function-or-method>`
- [x] The PEP8 style check passes: `git diff upstream/master -u -- "*.py" | flake8 --diff`
- [x] The html version looks good: `python doc/make.py --single <your-function-or-method>`
- [x] It has been proofread on language by another sprint participant

Please include the output of the validation script below between the "```" ticks:

```
(pandas_dev) root@kalih:~/pythonpanda/pandas# python scripts/validate_docstrings.py pandas.Period.dayofweek

################################################################################
##################### Docstring (pandas.Period.dayofweek)  #####################
################################################################################

Return the day of the week.

This attribute returns the day of the week on which the particular
starting date for the given period occurs with Monday=0, Sunday=6.

Returns
-------
Int
    Range of 0 to 6

See also
--------
Period.dayofyear
    Return the day of year.
Period.daysinmonth
    Return the days in that month.

Examples
--------
>>> period1 = pd.Period('2012-1-1 19:00', freq='H')
>>> period1
Period('2012-01-01 19:00', 'H')
>>> period1.dayofweek
6

>>> period2 = pd.Period('2013-1-9 11:00', freq='H')
>>> period2
Period('2013-01-09 11:00', 'H')
>>> period2.dayofweek
2

################################################################################
################################## Validation ##################################
################################################################################

Docstring for "pandas.Period.dayofweek" correct. :)
```

If the validation script still gives errors, but you think there is a good reason
to deviate in this case (and there are certainly such cases), please state this
explicitly.
