Exact value of the summation from k = 1 to infinity of 3/(4^k)

[Metronome]

In particular, I'm trying to figure out what solution process I used on the exam. Unfortunately, my handwriting and organizational skills were extremely poor.

Can anyone figure out how I got the correct answer of 1?

 

[ksdhart2]

Wow, you, uh, you weren't kidding about that. To me, this just looks like a harried mess of scribbles. I can't follow anything that's going on here, nor make heads or tails of what it might mean. It sorta almost looks like you might have been trying to set up a telescoping series and cancel the like terms... but I dunno. In any case, your best at this point is almost certainly to start over and work through it again, writing out the steps in a much clearer fashion. For instance, you might start:

\(\displaystyle \displaystyle \sum \limits_{k=1}^{\infty} \dfrac{3}{4^k} = 3 \cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{4^k} = 3 \cdot \sum \limits_{k=1}^{\infty} \left( \dfrac{1}{4} \right)^k\)

Where does this take you? Try finishing up from here.

 

[Metronome]

I can apply the geometric series definition from there; 3 * (1/4)/(1 - (1/4)) = 1.

Still curious about how the telescoping series process could be applied to this problem.

 

[Dr.Peterson]

I can apply the geometric series definition from there; 3 * (1/4)/(1 - (1/4)) = 1.

Still curious about how the telescoping series process could be applied to this problem.

I think what you actually did was effectively to derive the formula for the geometric series, by subtracting 1/4 of the sum from the sum itself.

This looks superficially like telescoping, but is not, unless you think of it as making a telescoping series out of the given one. A telescoping series is one in which parts of successive terms of a series cancel; here you are canceling corresponding terms in two series.