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 $3.5$

Calculus is not required to evaluate the sum.  Let $$f(z) = \sum_{k=1}^\infty (k+1)^2 z^k.$$  Then $$z f(z) = \sum_{k=1}^\infty (k+1)^2 z^{k+1} = -z + \sum_{k=1}^\infty k^2 z^k$$ hence $$f(z) - zf(z) = \sum_{k=1}^\infty (k+1)^2 z^k + z - \sum_{k=1}^\infty k^2 z^k = z + \sum_{k=1}^\infty (2k+1) z^k.$$  Now let $$g(z) = \sum_{k=1}^\infty (2k+1) z^k.$$  Then using the same technique, $$g(z) - z g(z) = \sum_{k=1}^\infty (2k+1)z^k - \sum_{k=2}^\infty (2k-1)z^k = z + 2\sum_{k=1}^\infty  z^k = z + \frac{2z}{1-z}.$$  Therefore, $$g(z) = \frac{z(3-z)}{(1-z)^2},$$ and $$f(z) = \frac{z(4-3z+z^2)}{(1-z)^3}.$$  All that remains is to select $z = 1/3$ to obtain the value of the desired sum.
