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Find the sum of the series $$\sum_{k=0}^\infty \frac{(-1)^k}{5^k}$$

I'm wondering if this is divergent since (-1)^k is divergent as per the rules of geometric series where $abs(x) \geq 1$ then $\sum_{k=0}^\infty x^k$ diverges.

Since we know that $$\sum_{k=0}^\infty \frac{(-1)^k}{5^k}=$$

$$\sum_{k=0}^\infty (-1)^k\frac{1}{5^k}$$ and $(-1)^k$ is a divergent then the rest of thte series is divergent? Is that logic right?

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    $\begingroup$ Look up the alternating series test. $\endgroup$ – Sean Nemetz Oct 20 '19 at 20:45
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    $\begingroup$ $\sum_{k=0}^\infty \frac{(-1)^k}{5^k}=\sum_{k=0}^{\infty}(\frac{1}{-5})^k=\frac{1}{1--\frac{1}{5}}=\frac{5}{6}$ by geometric series. $\endgroup$ – Locally unskillful Oct 20 '19 at 20:52
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This is a geometric series: $$\sum_{n=0}^{\infty}{r^n}=\dfrac{1}{1-r}$$ when $|r|<1$.

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  • $\begingroup$ With alternating signs, it's rather $\dfrac1{1+r}$. $\endgroup$ – Bernard Oct 20 '19 at 20:57
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    $\begingroup$ @Bernard No it's not, it's just that $r$ is negative. $\endgroup$ – kccu Oct 20 '19 at 21:02
  • $\begingroup$ It is correct. He just have to write it as $\sum_{k=0}^{\infty}\left(-\dfrac15\right)^k$. In this fashion $r=-\dfrac15$. $\endgroup$ – DINEDINE Oct 20 '19 at 21:03
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    $\begingroup$ @Bernard why? that requires remembering an extra rule while the one used above is incredibly general and covers both cases by itself. $\endgroup$ – JMoravitz Oct 20 '19 at 21:15
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    $\begingroup$ Daily vote limit reached so I cannot upvote your comment @JMoravitz, but I agree. There is no need to remember anything besides the formula for $\frac1{1-r}$. Remembering a separate formula for $\frac1{1+r}$ is just unnecessary. $\endgroup$ – Math1000 Oct 20 '19 at 21:22

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