# 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.

• Is this homework? Oct 6, 2010 at 21:29
• Yes, but it's the end of a problem for a Financial Mathematics course dealing with internal rate of return. I can write that the answer is 3.56592, but I want to know how to solve this problem.
– s c
Oct 6, 2010 at 21:32
• $\sum_{n=1}^{t}$ does not make much sense when $t$ is fractional. Perhaps the question is phrased differently? Oct 6, 2010 at 21:40
• @Moron: The sum $S$ is a function of the number of terms, but we can "continue" $S(t)$ to the reals. Oct 7, 2010 at 10:56
• @Moron: In my opinion, only the context establishes its relevancy or not. If one is asked to interpolate between number of capitalization periods (e.g. to analyze a scenario with different interest rates) it makes sense. Or when one studies the sensitivity of $S(t)$ to deviation from the estimated life ($t$ periods, months, years,etc.). See e.g. chapter on Sensitivity Analysis in Engineering Economics by Riggs, Bedforth and Randhava. Oct 7, 2010 at 14:51

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:

$$S=u_{1}+u_{2}+u_{3}+\ldots +u_{t}$$

$$rS=ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{t-1}+ru_{t}$$

$$u_{k}=ru_{k-1}=u_{1}r^{k-1}$$

$$S-rS=\left( u_{1}+u_{2}+u_{3}+\ldots +u_{t}\right) -\left( ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{t-1}+ru_{t}\right)$$

$$(1-r)S=u_{1}-ru_{t}$$

$$S=\dfrac{u_{1}-ru_{t}}{1-r}=\dfrac{u_{1}-u_{1}r^{t}}{1-r}$$

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", $\rm\ 1/(a-1) = 1/0.08 = 100/8 = 12.5$

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.)