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Find whether the series diverges and its sum: $$\sum_{n = 1}^\infty (-1)^{n+1} \frac{3}{5^n}.$$

I found that the series converges using the Alternating Series test because the absolute value of each $n$ decreases while the value of $n$ increases. Then I took the limit as $x$ approaches infinity of $3/5^x$ which is $0$.

But I am not sure how to go about finding the sum at this point. This is a practice exam, so the solution is more important then the actual answer.

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Hint: it's a Geometric series. –  David Mitra Dec 8 '12 at 21:02
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Notice that$$(-1)^{n+1}\frac{3}{5^n}=-3\frac{(-1)^{n}}{5^{n}}=-3\left(\frac{-1}{5}\right)^{n}$$ Since $\sum_{k=1}^{\infty}ar^{k}=\frac{ar}{1-r}$ (iff $|r|<1$), $$\sum_{n=1}^{\infty}-3\left(\frac{-1}{5}\right)^{n}=\frac{-3\cdot\frac{-1}{5}}{1-\frac{-1}{5}}=\frac{\frac{3}{5}}{\frac{6}{5}}=\frac{1}{2}$$ and the sum converges because $\left|\frac{-1}{5}\right|=\frac{1}{5}<1$

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Hint: Use the following

$$ \sum_{k=0}^{\infty}(-1)^kx^k=\frac{1}{1+x}. $$

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Your series is an example of a geometric series. The first term is $a=3/5$, while each subsequent term is found by multiplying the previous term by the common ratio $r=-1/5$.

There is a well known formula for the sum to infinity of a geometric series with $|r| < 1$, namely:

$$S_{\infty} = \frac{a}{1-r} \, . $$

In your case, $a=3/5$ and $r=-1/5$, and so it follows that:

$$S_{\infty} = \frac{3/5}{1+1/5} = \frac{1}{2} \, . $$

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