Perpetuity Valuation Question

SUM(i1000/(1.1^i)) for i = 1,2,....,10 1000SUM(i/(1,1^i)) and there's a math formula for this but i can`t remember it (i think it was an arithmetic or geometric sum, last time i saw that was in my first year in college)

Assuming the cash flow is at the end of each year and the discount rate you're given is on an annual compounding basis, it's 1000$ times the sum from i = 1 to infinity of i/(1.1^i)

This infinite series is known as Gabriel's Staircase. The formula for it is r/(1-r)^2 where r is 1/(1.1)

I'm aware of at least two derivations for that formula, neither of which take more than 2 lines of algebra

The numerical answer you get is 110 000 $

a geometric sum. However, you can write it as the derivative with respect to r of a geometric sum

Please post the workings nname. This is teasing me. Googled Gabriels Staircase but couldnt come up with the answer.

Thanks

Sorry

This is a pretty straight-forward question. Forget the $1,000 part (just start with $1, then $2, then $3, etc.). It is pretty straight-forward to write this as a double-sum. Let i be the annual interest rate for discounting (10% here)

sum_{j = 1 to Inf} sum_{k = j to Inf} (1+i)^(-k) = (1/i) * sum_{j = 0 to Inf} (1+i)^(-j) = (1+i) / i^2

Each equality follows via geometric sum (and a little algebra to clean up the expression).

This is a simple brainteaser, monkies

you have a perpetuity, in perpetuity

the first perpetuity one is worth:

1000/.10 = 10,000

this perpetuity, is happening every year. (y1, y2, y3, y4, so on and so forth) - so its value, can be calculated like a perpetuity as well:

10,000/.10 = 100,000

edit: interview questions like this are not meant to be done with calculators or excel, they exist to test if you can "breakdown" supposedly complex problems

aranaxon's approach is still what they were looking for though.

He just forgot to mention that this is a perpetuity in advance and add the first $10,000.

The ad hoc thought processes above are reasonable, but in order to get it right (i.e. not trying to guess your way to the correct solution, and not looking up some formula which you cannot do on the spot), turn the intuition into the double sum in my prior post. This is not hard to do, but perhaps a clearer explanation will elucidate:

One perpetuity starting at time 1: sum{k = 1 to Inf} 1/(1+i)^k A new perpetuity starting at time 2: sum{k = 2 to Inf} 1/(1+i)^k Yet another one at time 3: sum{k = 3 to Inf} 1/(1+i)^k ... We get a perpetuity like this for every time j = 1, 2, 3, ... Thus, the value is the double sum in prior post:

sum{j = 1 to Inf} sum{k = j to Inf} 1/(1+i)^k = (1+i) / i^2

I argued why this equality holds previously (two separate geometric series). Here: i = 10%, so 1.1 / 0.1^2 = 1,100 Since there is a notional amount of $1,000, we get a final answer of ($1,000)*(1,100) = $110,000. This is the way to do it. That is, a combination of the right intuition mentioned by others, and just a tiny bit of math (only geometric sums).

That formula is a classic, and is well-known to any math undergraduate or participant in high school math competitions.

Even if you don't know it, deriving it is EXACTLY what aranaxon did (though, as I said, he made a mistake in doing so).

nnamehere, I understand what you are saying, but in my experience, strong math people rarely memorize formulas, but rather re-derive relatively simple formulations such as this. I say this, because if I change the problem just a little bit, then this "formula" is no longer useful. Consider a perpetuity which pays 1, 3, 6, 10, 15, 21, 28, 36, 46, 57, ... If we discount at rate interest rate i, what is the value? Answer: (1+i)^2 / i^3. Anyway, we have probably beaten this discussion to death.

I'm not disagreeing with you that you should be able to reconstruct it from scratch.

Indeed, doing so (either of the ways detailed so far) gives you the method to find the value of ANY polynomial payoff (not just constant or linearly increasing) discounted at a constant rate.