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

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?

• Look up the alternating series test. – Sean Nemetz Oct 20 '19 at 20:45
• $\sum_{k=0}^\infty \frac{(-1)^k}{5^k}=\sum_{k=0}^{\infty}(\frac{1}{-5})^k=\frac{1}{1--\frac{1}{5}}=\frac{5}{6}$ by geometric series. – Locally unskillful Oct 20 '19 at 20:52

This is a geometric series: $$\sum_{n=0}^{\infty}{r^n}=\dfrac{1}{1-r}$$ when $$|r|<1$$.
• With alternating signs, it's rather $\dfrac1{1+r}$. – Bernard Oct 20 '19 at 20:57
• @Bernard No it's not, it's just that $r$ is negative. – kccu Oct 20 '19 at 21:02
• It is correct. He just have to write it as $\sum_{k=0}^{\infty}\left(-\dfrac15\right)^k$. In this fashion $r=-\dfrac15$. – DINEDINE Oct 20 '19 at 21:03
• 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. – Math1000 Oct 20 '19 at 21:22