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

Sep 28, 2014

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... ${S}_{10} = {\sum}_{n = 1}^{10} {\left(0.6\right)}^{n - 1}$
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:

${S}_{10} - 0.6 {S}_{10} = {\left(0.6\right)}^{0} - {\left(0.6\right)}^{10}$ all the middle terms cancel!
${S}_{10} \left(1 - 0.6\right) = 1 - {0.6}^{10}$ , so by dividing by 0.4 = 2/5:
${S}_{10} = \frac{1 - {0.6}^{10}}{0.4} = 2.5 \left(1 - {0.6}^{10}\right)$
If you calculate the power exactly you get
${S}_{10} = 2.5 \left(1 - 0.00604662\right) = 2.5 \left(0.993953\right) = 2.48488 .$

The general formula ${S}_{k} = {\sum}_{n = 1}^{k} a {r}^{n - 1} = \frac{a \left(1 - {r}^{k}\right)}{1 - r}$
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./