What is the sum of this infinite geometric series?

[aces9113]

Homework Statement

∞
∑ ( 60^(1/(n+3)) − 60^(1/(n+4)) )
n = 0

Homework Equations

I believe this is a geometric series so the sum would equal a/(1-r)

The Attempt at a Solution

I tried to view it as a geometric series but i had trouble finding a ratio, especially what i thought was r (60) would make the series diverge

 

[Dick]

You are right that it's not geometric. You aren't going to have much luck actually finding a sum. Do you just want to show it converges?

 

[grey_earl]

Ever heard of the term telescopic sum? Try writing the sum up to n = 1, then n = 2, then for all n up to n = 3, and you will see what I mean.

(BTW: The answer is 60^(1/3) - 1.)

 

[aces9113]

yes but I'm not entirely sure how to do them. when I do that, I am left with 60^(1/3) and then 60^(1/infinity+3) - 60^(1/infinity+4) which would end up being 60^(1/3) since the others cancel but this is wrong. what am i doing wrong?

 

[Dick]

yes but I'm not entirely sure how to do them. when I do that, I am left with 60^(1/3) and then 60^(1/infinity+3) - 60^(1/infinity+4) which would end up being 60^(1/3) since the others cancel but this is wrong. what am i doing wrong?

Oh yeah, it does telescope in a manner of speaking. Thanks grey_earl. The terms in the split up series don't go to zero. You need to think of it as a limit of partial sums.

 

[grey_earl]

Exactly. Calculate first $$\sum_{n=0}^N (60^{1/(n+3)} -60^{1/(n+4)})$$, and then take the limit $$N \rightarrow \infty$$.