Evaluating $\sum _{n=2}^{\infty } \frac{(-5)^n}{8^{2 n}}$ making use of geometric series

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?

• No, it's not. But note $(-5)^n/ 8^{2n}= (-5/8^2)^n$. – David Mitra Jan 16 '15 at 20:24
• it is a good idea to write out the first few terms. – abel Jan 16 '15 at 20:25

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 $$\sum\limits_{n=2}^{\infty } \left(-\frac{5}{64}\right)^n=\frac{\frac{25}{4096}}{1-\left(-\frac{5}{64}\right)}=\frac{25}{4416}$$

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

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

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.

• 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? – bjd2385 Jan 16 '15 at 20:29
• 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 $c$. – John Hughes Jan 16 '15 at 20:34

Hint: $$\sum_{n=2}^{\infty} \left(\frac{-5}{64}\right)^n$$