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Evaluate $$\sum _{n=2}^{\infty}\frac{(-5)^n}{8^{2n}}$$ using geometric series.

I thought it would be possible to split this series such that we have

$$\sum _{n=2}^{\infty } (-5)^n \cdot \sum _{n=2}^{\infty } \left(\frac{1}{8}\right)^{2 n}$$

However, I am not sure that this is actually possible and I also see that the first sum does not converge, so even if it was possible I am not able to solve it. Could someone walk me through the steps?

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  • $\begingroup$ No, it's not. But note $(-5)^n/ 8^{2n}= (-5/8^2)^n$. $\endgroup$ – David Mitra Jan 16 '15 at 20:24
  • $\begingroup$ it is a good idea to write out the first few terms. $\endgroup$ – abel Jan 16 '15 at 20:25
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First note that $$ \sum\limits_{n=2}^{\infty } \frac{(-5)^n}{8^{2 n}}= \sum\limits_{n=2}^{\infty } \left(-\frac{5}{64}\right)^n $$

Now let's look at the first two terms of the sum $$ \left(-\frac{5}{64}\right)^2+ \left(-\frac{5}{64}\right)^3+\dots $$ $$ =\left(\frac{5}{64}\right)^2- \left(\frac{5}{64}\right)^3+\dots $$

So now we know that $$ a= \left(\frac{5}{64}\right)^2=\frac{25}{4096} $$ And $$ r= -\frac{5}{64} $$ Therefore $$ \sum\limits_{n=2}^{\infty } \left(-\frac{5}{64}\right)^n=\frac{\frac{25}{4096}}{1-\left(-\frac{5}{64}\right)}=\frac{25}{4416} $$

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HINT: It is $\sum_{n=2}^{\infty}\left(\frac{-5}{64}\right)^n$.

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Try $$\displaystyle \sum _{n=2}^{\infty } \left(-\frac{5}{64}\right)^n = \frac{25}{4096}\sum _{n=0}^{\infty } \left(-\frac{5}{64}\right)^n$$ as a typical geometric series

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That's not possible, no. But since $8^{2n} = 64^n$, you can rewrite your series as $$ \sum \left(\frac{-5}{64}\right)^n, $$ which should get you on your way.

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  • $\begingroup$ Could you not evaluate it first using the fact that $$\sum\limits_{n=0}^\infty\left(\frac{-5}{64}\right)^n=\frac{1}{1-\left(\frac{-5}{64}\right)}$$ and the subtracting the first two terms from that fraction, since $n=0$ and $n=1$ are not included in the bounds, then simplify? $\endgroup$ – bjd2385 Jan 16 '15 at 20:29
  • $\begingroup$ Yes, that's exactly right. Alternatively, you could let $a = -5/64$, and call your sum $S$, and note that $\frac{S}{a^2}$ is the sum from $0$ to infinity, so that $S$ must be $a^2$ times that sum. In general, $\sum_0 c r^n = \frac{c}{1 - r}$ lets you handle sums starting at nonzero values (like the "2" in your problem) by factoring out some power of $r$ and including that in $c$. $\endgroup$ – John Hughes Jan 16 '15 at 20:34
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Hint: $$\sum_{n=2}^{\infty} \left(\frac{-5}{64}\right)^n$$

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