<s>[INST] 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!
 [/INST] depend on the geometric series , we can get the following to compute the sum
$${\frac {x^2}{1-x^2}}=\sum _{n=1}^{\infty }x^{2n}\quad {\text{ for }}|x|<1\!$$
$${\displaystyle {\frac {x^2}{(1-x^2)^{2}}}=\sum _{n=1}^{\infty }nx^{2n}\quad {\text{ for }}|x|<1\!}$$
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