<s>[INST] Explicit sum of $\sum_{n=0}^{\infty} \frac{6+3^n}{6^{n+2}}$ I want to find the sum of the infinite series $$\sum_{n=0}^{\infty} \frac{6+3^n}{6^{n+2}}$$
So far I have managed to simplify the expression to $$\sum_{n=0}^{\infty} \left(\frac {1}{6^{n+1}} + \frac{1}{9\cdot2^{n+2}}\right)$$.
The sum clearly converges as both expressions are equal to $0$ as $\lim_{n\to\infty}$. Any tips as to how I may find an explicit expression for the sum of this series?
thanks in advance
 [/INST] As mentioned,it is geometric series $$\sum _{ n=0 }^{ \infty  } \left( \frac { 1 }{ 6^{ n+1 } } +\frac { 1 }{ 9\cdot 2^{ n+2 } }  \right) =\frac { 1 }{ 6 } \sum _{ n=0 }^{ \infty  } \frac { 1 }{ 6^{ n } } +\frac { 1 }{ 36 } \sum _{ n=0 }^{ \infty  } \frac { 1 }{ 2^{ n } } =\\ =\frac { 1 }{ 6 } \left( 1+\frac { 1 }{ 6 } +\frac { 1 }{ { 6 }^{ 2 } } +... \right) +\frac { 1 }{ 36 } \left( 1+\frac { 1 }{ 2 } +\frac { 1 }{ { 2 }^{ 2 } } +... \right) =\frac { 1 }{ 6 } \frac { 1 }{ 1-\frac { 1 }{ 6 }  } +\frac { 1 }{ 36 } \frac { 1 }{ 1-\frac { 1 }{ 2 }  } =\\ =\frac { 1 }{ 5 } +\frac { 1 }{ 18 } =\frac { 23 }{ 90 } $$
</s>