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mathy1991
12-12-2011, 09:03 AM
Hi, just wondering can anyone tell em how to sum this term from k=1 to infinity?

I used to know how to do it but I forgot, it's something to do with identifying a partial sum,

it is the sum of (e^(-0.02k))

Dason
12-12-2011, 09:34 AM
Let X = \sum_{k = 1}^\infty \exp(-.02k)

Let's look at what this sum sort of looks like...
X = \exp(-.02) + \exp(-.02*2) + \exp(-.02*3) + \ldots

Now what happens if we pull that first term out?

X = \exp(-.02) * [\exp(-.02(1-1) + \exp(-.02(2-1)) +\exp(-.02(3-1)) + \ldots]
X = \exp(-.02)*[1 + \exp(-.02) + \exp(-.02*2) + \exp(-.02*3) + \ldots]

Then notice that inside that parenthesis we have the original series we're trying to figure out. This gives us an equation of the form X = K(1 + X) and what you're trying to solve for is X.

Hopefully that hint should get you there. We didn't really need to write it out what the sum actually looked like and could have instead just worked using summation notation but I think this way helps you visualize what is going on better.

mathy1991
12-12-2011, 09:39 AM
Let X = \sum_{k = 1}^\infty \exp(-.02k)

Let's look at what this sum sort of looks like...
X = \exp(-.02) + \exp(-.02*2) + \exp(-.02*3) + \ldots

Now what happens if we pull that first term out?

X = \exp(-.02) * [\exp(-.02(1-1) + \exp(-.02(2-1)) +\exp(-.02(3-1)) + \ldots]
X = \exp(-.02)*[1 + \exp(-.02) + \exp(-.02*2) + \exp(-.02*3) + \ldots]

Then notice that inside that parenthesis we have the original series we're trying to figure out. This gives us an equation of the form X = K(1 + X) and what you're trying to solve for is X.

Hopefully that hint should get you there. We didn't really need to write it out what the sum actually looked like and could have instead just worked using summation notation but I think this way helps you visualize what is going on better.
Oh right thanks, so is the answer e^-0.02*(1/(1-e^-0.02))?

SmoothJohn
12-12-2011, 09:41 AM
Now's a good time to investigate. Check the partial sum (maybe using excel) and compare it to your answer.

Dason
12-12-2011, 09:43 AM
Looks fine to me. One way to check if this seems reasonable is just to have something compute the partial sum up to an arbitrarily large number for you. Using R this is trivial.

> k <- 1:10000
> sum(exp(-.02*k))
> # spits out an answer
> exp(-.02)/(1 - exp(-.02))