<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] This is not a geometric series on its own, it's the difference of two of them
$$\sum_{n\ge 3} \left({2\over 3}\right)^n-\sum_{n\ge 3}\left({1\over 3}\right)^n = {8/27\over 1-2/3}-{1/27\over 1-1/3}={8\over 9}-{1\over 18}={5\over 6}.$$
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