I want my book to be away from this table
Octave, 41 40 33 bytes
1 byte saved thanks to @Dennis

@(n)find(cumsum(.5./(1:9^n))&gt;n,1)

Try it online!
Explanation
This uses the fact that harmonic numbers can be lower-bounded by a logarithmic function.
Also, the &gt;= comparison can be replaced by &gt; because harmonic numbers cannot be even integers (thanks, @Dennis!).
@(n)                                   % Anonymous function of n
                     1:9^n             % Range [1 2 ... 9^n]
                .5./(     )            % Divide .5 by each entry
         cumsum(           )           % Cumulative sum
                            &gt;n         % Is each entry greater than n?
    find(                     ,1)      % Index of first true entry