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I need to find the sum of the following series:

$$\sum_{i=0}^\infty {3 * (-1)^{n+1} \over 2^{n}}$$

I started to simplify this:

$$3 *\sum_{i=0}^\infty {(-1)^{n+1} \over 2^{n}}$$

And then I stuck. any idea how to continue?


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up vote 5 down vote accepted

Recall that

$$\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$$

Now substutute $x=-y$ and get

$$\sum_{n=0}^{\infty} (-y)^n = \frac{1}{1+y}$$

Note that $(-y)^n = (-1)^n y^n$. Substitute $y=1/2$.

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This series is like this : $$3\left(-1+\frac 12-\frac {1}{2^2}+\frac {1}{2^3}+\cdots\infty\right)$$

this infinite geometric series where $a=-1,r=-\dfrac 12$

sum of infinite geometric series $S_{\infty}=\dfrac {a}{1-r}$ where $|r|<1$

$$S_{\infty}=3\dfrac {-1}{1-(-\frac 12)}$$ $$S_{\infty}=3\dfrac {-1}{1+\frac 12}$$ $$S_{\infty}=-2$$

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