<s>[INST] Find the sum of $\displaystyle\sum_{n=3}^\infty \frac{2^n-1}{3^n}$ My work:$$\sum_{n=3}^\infty \frac{2^n-1}{3^n}$$
$$a_1 = \frac{2^3-1}{3^3}$$
$$a_1 = \frac{7}{27}, r=\frac{2}{3}$$
$$S_N=\frac{\frac{7}{27}}{1-\frac{2}{3}} = \frac{7}{9}$$
The correct answer is $\frac{5}{6}$
 [/INST] Both geometric series are convergent, so
$$S=\sum_{n=3}^{+\infty}(\frac{2}{3})^n-\sum_{n=3}^{+\infty}\frac{1}{3^n}$$
$$\frac{2^3}{3^3}\frac{1}{1-\frac{2}{3}}-\frac{1}{3^3}\frac{1}{1-\frac{1}{3}}$$
$$=\frac{8}{9}-\frac{1}{18}=\frac{5}{6}$$
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