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

1 Answer
dansmath
Sep 28, 2014

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./

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