Evaluate $\frac{1+3}{3}+\frac{1+3+5}{3^2}+\frac{1+3+5+7}{3^3}+\cdots$ It can be rewritten as
$$S = \frac{2^2}{3}+\frac{3^2}{3^2}+\frac{4^2}{3^3}+\cdots$$
When $k$ approaches infinity, the term $\frac{(k+1)^2}{3^k}$ approaches zero. But, i wonder if it can be used to determine the value of $S$. Any idea?
Note: By using a programming language, i found that the value of $S$ is $3.5$

Hint: Let $f(t)=\sum t^{k}$. Then $tf'(t)=\sum kt^{k}$  and $t^{2}f''(t)=\sum (k^{2}-k)t^{k}$. This gives $\sum kt^{k}$ and $\sum k^{2}t^{k}$ in terms of $f,f'$ and $f''$. Put $t=1/3$ and use the fact that $(k+1)^{2}=k^{2}+2k+1$.
