Expanding your problem:
$ \sum\limits_{k = 0}^\infty {\frac{{k + 1}}{{3^k }}} = \frac{1}{3^0} + \frac{2}{3^1} + \frac{3}{3^2} + \frac{4}{3^3} + \dots $
$ = 1 + \left (\frac{1}{3} + \frac{1}{3} \right) + \left(\frac{1}{3^2} + \frac{1}{3^2} + \frac{1}{3^2} \right) + \left(\frac{1}{3^3} + \frac{1}{3^3}+ \frac{1}{3^3}+ \frac{1}{3^3}\right) + \dots$
This can be grouped into:
$ = \left(1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots\right)+ $ $ \left(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots\right)+ $ $ \left(\frac{1}{3^2} + \frac{1}{3^3} + \dots\right)+ $ $ \left(\frac{1}{3^3} + \dots\right) + \dots $
Using the fact that $ S = \sum_{n=0}^{\infty} \frac{1}{3^n} = \frac{3}{2}$: $ = \frac{3}{2} + $ $ \frac{3}{2} - (1) + $ $ \frac{3}{2} - \left(1 + \frac{1}{3} \right) + $ $ \frac{3}{2} - \left( 1 + \frac{1}{3} + \frac{1}{3^2} \right ) + \dots $
The partial sum $S_k$ is computed as: $S_k = \sum_{n=0}^k \frac{1}{3^n} = \frac{3}{2} - \frac{1}{2}\left(\frac{1}{3}\right)^k$
Hence, $ = \frac{3}{2} + \left(\frac{3}{2} - S_0 \right) + \left(\frac{3}{2} - S_1 \right) + \left(\frac{3}{2} - S_2 \right) \dots$ $ = \frac{3}{2} + \frac{1}{2} \left( 1 + \frac{1}{3} + \frac{1}{3^2} + \dots \right) $ $ = \frac{3}{2} + \frac{1}{2}S = \frac{3}{2} + \frac{1}{2} \frac{3}{2}$ $ = \mathbf{\frac{9}{4}}$