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The problem is to evaluate $$\sum_{k=1}^\infty 4\left(\frac{-1}{5}\right)^{4k}.$$ So I thought $r = (-1/5)$. And so it would converge since $(-1/5) < 1$. But i'm confused on how to get $a$ for the formula $a/(1-r)$. Is my $r$ wrong?

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2 Answers

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$$\sum_{k=1}^\infty 4\left(\frac{-1}{5}\right)^{4k}=4\sum_{k=1}^\infty \left(\frac{-1}{5}\right)^{4k}=4\sum_{k=1}^\infty \left(\frac{(-1)^4}{(5)^4}\right)^{k}=4\sum_{k=1}^\infty \left(\frac{1}{625}\right)^{k}$$

It's a geometric series. Because $r=\frac{1}{625}$ and therefore between $(-1, 1)$, the sum is $\frac{a_0}{1-r}$:

$$\therefore 4\left(\frac{\frac{1}{625}}{1-\frac{1}{625}}\right)=4\left(\frac{1}{624}\right)= \frac{1}{156}$$

Note that $a_0$ was $\frac{1}{625}$ because the sum is from $k=1$ and not $0$. Interestingly enough, if you did the classic $\frac{1}{1-r}$ (which is what it usually is if the first term is $1$/the sum starts from 0), you'd get $\frac{625}{624}$ which is higher than $\frac{1}{624}$ by one (of course, since the non-included $1$ is added).

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HintWrite out the first couple of terms. The first term is $a$ and the ratio between terms is $r$.

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