Sum of Series with negative exponents $
3= \sum_{n=1}^{t} \frac{1}{1.08^n}
$
I see that it is $3 = 1.08^{-t}(12.5 \times 1.08^t{-12.5})$ (from Wolfram Alpha, but I'm not sure how to get it. I tried solving as a geometric series, I had problems and didn't get the correct answer.
I see that $ t\approx3.56592$, which seems like it's correct, but I have no idea where the 12.5 and all that stuff came from. Unfortunately my calculus book doesn't help much, as it is mainly focus on infinite series.
 A: I would solve for $t$
$$S(t)=\displaystyle\sum_{n=1}^{t}\dfrac{1}{1.08^{n}}=3.$$
The sum of a geometric progression with first term $u_{1}$ and ratio $r$ is
given $(\ast)$ by
$$S(t)=u_{1}\times \dfrac{1-r^{t}}{1-r}.$$
In this case
$$\dfrac{1}{1.08}\times \dfrac{1-\left( \dfrac{1}{1.08}\right) ^{t}}{1-\dfrac{1}{%
1.08}}=3$$
or
$$\dfrac{1-\left( \dfrac{1}{1.08}\right) ^{t}}{0.08}=3\iff \left( \dfrac{1}{1.08}%
\right) ^{t}=0.76.$$
Applying logarithms, we obtain the "time" $t$
$$t\log \left( \dfrac{1}{1.08}\right) =\log 0.76,$$
$$t=\dfrac{\log 0.76}{-\log 1.08}\approx 3.5659,$$
which means that we need more than $3$ periods and less than $4$, at an interest rate of $8\%$, coumpounded per period, to get a total of $3$ currency units.
$(\ast)$ Derivation:
$$S=u_{1}+u_{2}+u_{3}+\ldots +u_{t}$$
$$rS=ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{t-1}+ru_{t}$$
$$u_{k}=ru_{k-1}=u_{1}r^{k-1}$$ 
$$S-rS=\left( u_{1}+u_{2}+u_{3}+\ldots +u_{t}\right) -\left(
ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{t-1}+ru_{t}\right) $$
$$(1-r)S=u_{1}-ru_{t}$$
$$S=\dfrac{u_{1}-ru_{t}}{1-r}=\dfrac{u_{1}-u_{1}r^{t}}{1-r}$$
A: HINT$\ $ Put $\rm\ a = 1.08\:$. Multiply both sides by $\rm\ a^t\ $ to get$\:$ RHS $\rm\: = \ \sum_{k=0}^{t-1}\ a^k$
As for "where the $12.5$ came from", $\rm\ 1/(a-1) = 1/0.08 = 100/8 = 12.5$
A: Simply suppose that $r = \frac{1}{1.08}$, or $r = a^{-k} = \frac{1}{a^k}$ and solve in the usual manner, as you would if the exponent was positive.
Example
$$ \sum \limits_{i=0}^{k-1} 9^{-i} = \sum \limits_{i=0}^{k-1} \frac{1}{9^i} = \frac{1- \left( 1/9 \right)^k}{1- \left( 1/9 \right)} = \frac{1 - \left( 1 / 9 \right)^k}{8/9} = \frac{9}{8} \left( 1 - \frac{1}{9^{k}} \right) $$
(Do not be confused with the harmonic series which does not have an exponent in its term.)
