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 $3.5$
 A: 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.
A: 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 $(k+1)^{2}=k^{2}+2k+1$.
A: Let $\dfrac{(k+1)^2}{3^k}=f(k)-f(k-1)$ where $f(n)=\dfrac{an^2+bn+c}{3^n}$
$\implies\dfrac{(k+1)^2}{3^k}=\dfrac{(ak^2+kb+c)-3(a(k-1)^2+b(k-1)+c)}{3^k}$
$\implies k^2+2k+1=-2ak^2+k(b+6a-3b)+c-3(a-b+c)$
Compare the coefficients of $k^2$ to find $$-2a=1$$
and by the coefficients of $k,$ $$6a-2b=2\iff b=?$$
and by the constants, $$1=3b-3a-2c\iff c=?$$
Now use Telescoping series
Of course we need to establish $$\lim_{n\to\infty}f(n)=0$$
A: $$\left(\frac13+\frac1{3^2}+\frac1{3^3}+\cdots \right)\left(1+3+\frac53+\frac7{3^2}+\cdots\right)$$
\begin{aligned}=\left(\frac12\right)\left(1+3+  \frac33+\frac3{3^2}+...\\
+\frac23+\frac2{3^2}+\cdots\\+\frac2{3^2}+\cdots\\+\cdots\right)\\
=\frac12\left(4+\frac32\\+1\\+\frac13\\+\cdots\right)\\
=\frac12\left(4+\frac32+\frac32\right)\end{aligned}
A: 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
$$ \frac{1}{3}+\frac{1+3}{3^2}+\frac{1+3+5}{3^3}+\frac{1+3+5+7}{3^4}+\ldots = \color{red}{\frac{7}{2}}.\tag{5} $$
