# Infinite series related interest question

1. Mar 22, 2016

### Zacarias Nason

1. The problem statement, all variables and given/known data

"A dollar due to be paid to you at the end of n months, with the same interest rate as in Problem 13, is worth only (1.005)^{-n} dollars now (because that is what will amount to $1 after n months). How much must you deposit now in order to be able to withdraw$10 a month forever (starting one month from now)? (Assume the interest rate remains the same forever.)"

2. Relevant equations

$$S_n = \frac{a(1-r^n)}{1-r}$$ $$S = \frac{a}{1-r}$$
The interest rate of Problem 13 is 6% compounded monthly, or r = 1.005.

3. The attempt at a solution

I'm almost totally unclear as to how we're supposed to solve this. "Forever" would imply to me that we're not dealing with a finite sum so the top formula isn't relevant, but I'm not sure how this should be dealt with.

2. Mar 22, 2016

### SteamKing

Staff Emeritus
What you are dealing with here is an annuity, except you get to collect the same sum every month instead of every year, in perpetuity.

https://en.wikipedia.org/wiki/Annuity

3. Mar 22, 2016

You invest an amount X. At the end of ine month, it has earnt some interest. You withdraw $10. How much needs to remain in the investment? 4. Mar 22, 2016 ### Zacarias Nason I get what it's asking. That doesn't change that I don't understand how to mathematically approach it, though. 5. Mar 22, 2016 ### Ray Vickson You are mis-using the formulas. If r = 1.005, then $\sum_{n=0}^N a r^n = a (r^{N+1}-1)/(r-1)$ is the future value at time $N$. In that case it makes not much sense to take $N \to \infty$ because, infinitely far in the future you will have an infinite amount of money in the bank---exceeding by far the total wealth of all individuals, companies and governments on planet earth. For r = 1.005 the present value (which is what you need in this problem) would be $\sum_{n=0}^N a/r^n = a (1 - (1/r)^{N+1})/(1 - (1/r))$, and this makes perfectly good sense as $N \to \infty$, where it equals $a/(1 - (1/r)) = ra/(r-1)$. BTW: you can see right away that the formulas you were applying must be incorrect, because for $r > 1$ you get a negative value of $S$, as though putting money in the bank eventually makes your account overdrawn! 6. Mar 22, 2016 ### Zacarias Nason Maybe I should've put a question mark next to them so others clued in that both of those formulas were tentative as I already wasn't* sure if they could be used. I don't know if I misused them if I did virtually no math with them yet, but it's fair to say that since I didn't indicate otherwise it could seem that I thought they must be used. I just didn't want to leave that section blank. 7. Mar 22, 2016 ### haruspex My question was posed so as to show you how to approach it. There is a very simple and immediate answer to my question. Think about it again. 8. Mar 22, 2016 ### Zacarias Nason If I had to hazard a guess, the initial investment would have to at minimum be$2,000 since (0.005)(2,000)=10 or (1.005)(2,000)=2010; if you withdrew $10 monthly you'd indefinitely be receiving the same amount and never having the account balance or the actual interest paid increase, but that doesn't sound right to me, or at the very least there has to be a better/more analytical way to write this or do this. I also have no idea how Ray derived the two formulas he did below which resembled but were not the finite sum for a geometric series. 9. Mar 22, 2016 ### Ray Vickson I did not "derive" them; I just used standard algebraic formulas as, in fact, you did in your post #1!. The only difference is that where you used r = 1.005 you should have used $r' = 1/r = 1/1.005$, because you want present value instead of future value. Last edited: Mar 22, 2016 10. Mar 22, 2016 ### Zacarias Nason I wrote this: in post #4, I have no idea what you're talking about w.r.t. me using algebraic formulas in that post. I understand how to derive a formula for $$S_n = a + ar +ar^2 + ... + ar^{n-1} + ...$$, so I'm guessing you did something similar when you said "standard algebraic formulas"; I don't know what you did because apparently$2,000 is the answer and if asking where those present and future value formulas came about will get, "I used algebra" in response, it isn't worth pressing. Thanks.

11. Mar 22, 2016

### Ray Vickson

Sorry, that was a typo: I meant post #1, and have edited the previous message to reflect that fact.

By "standard algebraic formulas" I mean the formulas you actually used yourself in post #1, when you wrote value for $S$ in terms of $a$ and $r$. However, there are alternative, slicker, algebraic methods that lead more easily and more quickly to the answer, and you seemed to be applying those faster methods in your post #8. You will arrive at the same answer either way.

BTW: you will get slightly different answers depending on when the first withdrawal is made. If you make your first withdrawal at time 0 (that is, you put in $\S$ and immediately withdraw $\10$) you get a higher value of $S$ than if your first withdrawal occurs at time 1, the end of the first month.

Last edited: Mar 22, 2016
12. Mar 22, 2016

You have the formula in post 8 already, where you checked the answer $2000. Solve it for the$2000, and you can use the equation to "find" $2000. There is no need to calculate any finite or infinite sums - this problem would be much harder if the money was only supposed to last for N month. 13. Mar 24, 2016 ### Chestermiller ### Staff: Mentor Yes. I agree with the analysis in post #8. If you're earning half a percent a month, then to take out$10 each month , you need a principal of \$2000.

14. Mar 24, 2016

### haruspex

Yes, that's exactly the argument I was hinting at. why apply a complicated formula for a general case when you have such a simple and direct route to the answer?