Compute $\sum\limits_{n=1}^\infty \frac{5+4n-1}{3^{2n+1}}$ I have troubles finding the limit of the following series: $\sum_{n=1}^\infty \frac{5+4n-1}{3^{2n+1}}$
So far I figured it'd easier to split the sum into:
$\sum_{n=1}^\infty \frac{5}{3^{2n+1}} \sum_{n=1}^\infty \frac{4n-1}{3^{2n+1}}$
= $\sum_{n=1}^\infty 5 \cdot\frac{1}{3^{2n+1}} +\sum_{n=1}^\infty 4n-1 \cdot \frac{1}{3^{2n+1}}$
And with $\sum_{n=1}^\infty \frac{1}{w^n} = \frac{1}{w-1}$ you get the following terms:
$5\cdot \frac{1}{3^{2n+1}-1} + 4n-1\cdot \frac{1}{3^{2n+1}-1}$
which is bascially a sequence, but im stuck right here.. help is very appreciated!
 A: depend on the geometric series , we can get the following to compute the sum
$${\frac {x^2}{1-x^2}}=\sum _{n=1}^{\infty }x^{2n}\quad {\text{ for }}|x|<1\!$$
$${\displaystyle {\frac {x^2}{(1-x^2)^{2}}}=\sum _{n=1}^{\infty }nx^{2n}\quad {\text{ for }}|x|<1\!}$$
A: Hint. We have that for $|x|<1$,
$$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}\quad\mbox{and}\quad \frac{d}{dx}\left(\sum_{n=0}^{\infty} x^n\right)=\sum_{n=0}^{\infty}n x^{n-1}=\frac{1}{(1-x)^2}.$$
Hence
$$\sum_{n=1}^\infty \frac{5+4n-1}{3^{2n+1}}=\sum_{n=1}^\infty \frac{4+4n}{3\cdot 9^{n}}=\frac{4}{3}\sum_{n=1}^\infty (1/9)^{n}+\frac{4}{27}\sum_{n=1}^\infty n(1/9)^{n-1}.$$
A: Let  $F_m=\dfrac{Am+B}{3^{2m+1}}$ and
 $\dfrac{4+4n}{3^{2n+1}}=F_{n-1}-F_n$
$\implies\dfrac{4+4n}{3^{2n+1}}=\dfrac{A(n-1)+B}{3^{2n-1}}-\dfrac{An+B}{3^{2n+1}}=\dfrac{9\{A(n-1)+B\}-(An+B)}{3^{2n+1}}$
$\implies4+4n=8B-9A+8An$
$\implies8A=4\iff A=\dfrac12,4=8B-9A\iff B=?$
Clearly, $$\sum_{n=1}^\infty\dfrac{4+4n}{3^{2n+1}}=\sum_{n=1}^\infty\left(F_{n-1}-F_n\right)=F_0-\lim_{n\to\infty}F_n$$
Now can you establish  $$\lim_{n\to\infty}F_n=0$$
