sum to infinity

#1
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))

Thanks in advance.
 

Dason

Ambassador to the humans
#2
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.
 
#3
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))?
 

Dason

Ambassador to the humans
#5
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.
Code:
> k <- 1:10000
> sum(exp(-.02*k))
> # spits out an answer
> exp(-.02)/(1 - exp(-.02))
> #spits out an answer
> # check if they're the same or at least really close.