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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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    $\begingroup$ Consider $4(\frac{1}{5^4})^k$. $\endgroup$
    – A. Wong
    Apr 3, 2014 at 20:44
  • $\begingroup$ Also, watch out, because the initial term in the series is not 4 (since the summation starts at $ \ k = 1 \ $ . $\endgroup$ Apr 3, 2014 at 21:10

2 Answers 2

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

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  • $\begingroup$ So I would just ignore the 4 out front? That was throwing me off $\endgroup$
    – Mahina
    Apr 3, 2014 at 20:48
  • $\begingroup$ @Mahina Ignore? Why? Write out the first three terms, what are they? $\endgroup$
    – gt6989b
    Apr 3, 2014 at 20:49
  • $\begingroup$ The way I was taught they are (-1/5)^4 which would be a, (-1/5)^5 which is ar, and (-1/5)^6 which is ar^2. Is that right? I'm not sure what to do with the 4 outside the parentheses. $\endgroup$
    – Mahina
    Apr 3, 2014 at 20:51
  • $\begingroup$ @Mahina You compute the terms by "plugging in" the values $k=1,2,3$ into the formula in the sum: $4\left(\frac{-1}{5}\right)^{4\cdot 1}, 4\left(\frac{-1}{5}\right)^{4\cdot 2}, 4\left(\frac{-1}{5}\right)^{4\cdot 3}$. Simplify these, and then you get first term to be $a$ and ratio to be $r$. Check that ratio of 2nd/1st and 3rd/2nd terms is the same. $\endgroup$
    – gt6989b
    Apr 3, 2014 at 20:53

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