Question

Determine whether the series converges or diverges, if it converges, find the sum. (problem below)

Answer 1

\[\sum_{n =1}^{\infty} \frac{ (2^{2n -3} - 3 }{ 6^{n+2}})\]

Answer 2

Split up that difference.
Extract the constant factors from the exponents as in rewrite 6^(n+2) as 6^2*6^n since the constant terms don't matter.
Note that 2^(2n) is the same as 4^n

Answer 3

\[\begin{align*}\sum_{n=1}^\infty\frac{2^{2n-3}-3}{6^{n+2}}&=\sum_{n=1}^\infty\frac{2^{2n-3}}{6^{n+2}}-\sum_{n=1}^\infty\frac{3}{6^{n+2}}\\
&=\frac{2^{-3}}{6^2}\sum_{n=1}^\infty\frac{2^{2n}}{6^{n}}-\frac{3}{6^2}\sum_{n=1}^\infty\frac{1}{6^{n}}\\
&=\frac{1}{8\cdot 36}\sum_{n=1}^\infty\frac{4^{n}}{6^{n}}-\frac{3}{36}\sum_{n=1}^\infty\frac{1}{6^{n}}\\
&=\frac{1}{288}\sum_{n=1}^\infty\left(\frac{4}{6}\right)^n-\frac{1}{12}\sum_{n=1}^\infty\frac{1}{6^{n}}\\
&=\frac{1}{288}\sum_{n=1}^\infty\left(\frac{2}{3}\right)^n-\frac{1}{12}\sum_{n=1}^\infty\frac{1}{6^{n}}
\end{align*}\]
What do you know about geometric series with |r|<1?