Question

How do you find the 10-th partial sum of the infinite series `sum_(n=1)^oo(0.6)^(n-1)` ?

Answer 1

Hi, thanks for asking.
You have a geometric series here with common ratio r = 0.6, and the partial sum `S_k` is what you get when you add the first k (a few or a million) terms.
You want the first ten terms... `S_10=sum_(n=1)^10 (0.6)^(n-1)`
`S_10 = (0.6)^0+(0.6)^1+…+(0.6)^9` multiply by 0.6:

`0.6 S_10=(0.6)^1+…+(0.6)^9+(0.6)^10` then subtract:

`S_10 - 0.6 S_10=(0.6)^0-(0.6)^10` all the middle terms cancel!
`S_10(1-0.6) = 1 - 0.6^10` , so by dividing by 0.4 = 2/5:
`S_10=(1-0.6^10)/0.4=2.5(1-0.6^10)`
If you calculate the power exactly you get
`S_10=2.5(1-0.00604662)=2.5(0.993953)=2.48488.`

The general formula `S_k=sum_(n=1)^k a r^(n-1)=(a(1-r^k))/(1-r)`
is also useful but we did better; we derived it from scratch!

As an extra brain rep, see if you can verify this formula using our subtraction method, and then maybe figure out the infinite sum.

\dansmath - made from scratch, every time./