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Determine the sum of the following series: $$\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} $$

My work: $$\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} = \sum_{n=1}^{\infty } \frac{-1}{7} (\frac{3}{7})^{n-1}$$ $$\sum_{n=1}^{\infty } ar^{n-1} = \frac{a}{1-r} = \frac{\frac{-1}{7}}{1-\frac{3}{7}} = -\frac{1}{4}$$

Why does this not work?

Sorry for the incorrect initial post!!!

Edit: -3 changed to (-3)

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If that answer is wrong, could the sum actually be $\sum_{n=1}^\infty \frac{(-3)^{n-1}}{7^n} $? – Maxim G. Jul 27 '14 at 23:28
What do you mean. It seems fine. – ReverseFlow Jul 27 '14 at 23:28
up vote 5 down vote accepted

If the sum is $(-3)^{n-1}$ then you cannot factor out a -1. Since $(-3)^{n-1} = (-1)^{n-1}(3)^{n-1}$

There would be no such thing as an alternating series if you could just move the negative sign out in front of the sum!

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$$\begin{align} \sum_{n=1}^{\infty } \frac{-3^{n-1}}{7^{n}} & = - \frac{1}{7} \sum_{n=1}^{\infty } (\frac{3}{7})^{n-1} \\ & = - \frac{1}{7} \frac{1}{1-\frac 3 7} \\ & = - \frac 1 4 \\[2ex] \sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} & = \frac{1}{7} \sum_{n=1}^{\infty } (-\frac{3}{7})^{n-1} \\ & = \frac{1}{7} \frac{1}{1+\frac 3 7} \\ & = \frac 1 {10} \end{align}$$

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$$-\frac{1}{3} \sum_{n=1}^{\infty} \frac{3^n}{7^n}=-\frac{1}{3} \sum_{n=1}^{\infty} \left ( \frac{3}{7} \right )^n=-\frac{1}{3} \sum_{n=0}^{\infty} \left ( \frac{3}{7} \right )^n+\frac{1}{3}=-\frac{1}{3} \frac{1}{1-\frac{3}{7}}+\frac{1}{3} \\ =-\frac{1}{3}\frac{7}{7-3}+\frac{1}{3}=-\frac{7}{12}+\frac{4}{12}=\frac{-1}{4}$$

EDIT: $$-\frac{1}{3} \sum_{n=1}^{\infty} \frac{(-3)^n}{7^n}=-\frac{1}{3} \sum_{n=0}^{\infty} \frac{(-3)^n}{7^n}+\frac{1}{3}=-\frac{1}{3} \frac{1}{1-\left (\frac{-3}{7} \right )}-\frac{1}{3}=-\frac{1}{3} \frac{1}{1+\frac{3}{7}}+\frac{1}{3}=-\frac{1}{3} \frac{7}{10}+\frac{1}{3}=-\frac{7}{30}+\frac{10}{30}=\frac{1}{10}$$

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