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 $\frac{5}{6}$
 A: This is not a geometric series on its own, it's the difference of two of them
$$\sum_{n\ge 3} \left({2\over 3}\right)^n-\sum_{n\ge 3}\left({1\over 3}\right)^n = {8/27\over 1-2/3}-{1/27\over 1-1/3}={8\over 9}-{1\over 18}={5\over 6}.$$
A: Both geometric series are convergent, so
$$S=\sum_{n=3}^{+\infty}(\frac{2}{3})^n-\sum_{n=3}^{+\infty}\frac{1}{3^n}$$
$$\frac{2^3}{3^3}\frac{1}{1-\frac{2}{3}}-\frac{1}{3^3}\frac{1}{1-\frac{1}{3}}$$
$$=\frac{8}{9}-\frac{1}{18}=\frac{5}{6}$$
A: Strictly for fun, note that
$$2^n-1=(2-1)(2^{n-1}+2^{n-2}+\cdots+2^{n-n})=\sum_{k=1}^n{2^n\over2^k}$$
so that
$$\sum_{n=1}^\infty{2^n-1\over3^n}=\sum_{n=1}^\infty\sum_{k=1}^n\left(2\over3\right)^n{1\over2^k}=\sum_{k=1}^\infty\sum_{n=k}^\infty{1\over2^k}\left(2\over3\right)^n=\sum_{k=1}^\infty{1\over2^k}{(2/3)^k\over1-(2/3)}=3\sum_{k=1}^\infty\left(1\over3\right)^k=3{1/3\over1-1/3}={3\over2}$$
and therefore
$$\sum_{n=3}^\infty{2^n-1\over3^n}=\left(\sum_{n=1}^\infty{2^n-1\over3^n}\right)-\left(2^1-1\over3^1\right)-\left(2^2-1\over3^2\right)={3\over2}-{1\over3}-{1\over3}={5\over6}$$
