The following elementary approach appears to be mechanically simpler than the approach found in the series chapter of some calculus texts.

Set $S\=0.$$\stackrel{\&conjugate0;}{45}$ = $0.454545\cdots$, so $100 S\=45.$$\stackrel{\&conjugate0;}{45}$ = 45.454545⋯.

Then, $100 S-S\=99 S\=45.0$ and $S\=45\/99$ so that the given repeating decimal is $3\+45\/99\=38\/11$.

The charm of this approach is that a simple algorithm can be extracted, namely, divide the repeating digits by ${10}^k-1$, where $k$ is the number of digits that are repeated.

Calculus texts, looking for an application of the geometric series, approach the problem as follows.

As in the simpler method, the final result is $3\+45\/99\=38\/11$.

Of course, the series advocate will claim that the given repeating decimal can immediately be represented as