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 $3.5$

If we assume $z\in(-1,1)$ and start with
$$ \sum_{n\geq 0} (2n+1) z^n = \frac{1+z}{(1-z)^2} \tag{1}$$
granted by stars&bars
$$ \frac{1}{(1-z)^{m+1}}=\sum_{n\geq 0}\binom{n+m}{m}z^n\tag{2} $$
by convolution it follows that
$$ \sum_{n\geq 0}\left(\sum_{k=0}^{n}(2k+1)\right) z^n = \frac{1+z}{(1-z)^3}\tag{3} $$
and by evaluating both sides of $(3)$ at $z=\frac{1}{3}$ we get
$$ \sum_{n\geq 0}\frac{(n+1)^2}{3^n} = \frac{9}{2}\tag{4} $$
so
$$ \frac{1}{3}+\frac{1+3}{3^2}+\frac{1+3+5}{3^3}+\frac{1+3+5+7}{3^4}+\ldots = \color{red}{\frac{7}{2}}.\tag{5} $$
