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May 7th, 2016, 08:02 PM  #1 
Member Joined: May 2015 From: Earth Posts: 64 Thanks: 0  Finding an approximation using series
So the problem I am struggling with is attached, and is 8b. I am to use a given power series to find the approximation of a an integral of said power series so that the error between the actual and approximated answer is less than .01  I got through most of the problem, to the point where I am left with the power series of 1/(1/3)^n, but i'm not sure what to do next. The answer key (also in the same attached file) tells me that since (1/3)^5 is less than .01 my answer is s sub 4, but I don't understand that. First, why am I setting (1/3)^n to be less than .01, since my series involves a negative number. And secondly, since it was proven that when n is 5 the error is less than .01, why is my answer s sub 4? Thanks. 
May 7th, 2016, 09:14 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,634 Thanks: 2620 Math Focus: Mainly analysis and algebra 
I assume that you mean that the series is $\sum \left(\frac13\right)^n$ since what you wrote doesn't converge. Because the series is an alternating series, the series of partial sums $s_n = \sum \limits_{k=1}^n a_k$ has the property $s_1 \gt s_3 \gt \ldots S \ldots \gt s_4 \gt s_2$ (or the reverse  depending on the sign of $s_1$) where $S$ is the limit of the partial sums. The important thing to notice then is that for any $n$ we have $$s_n \gt S \gt s_{n+1} \quad \text{or} \quad s_{n+1} \gt S \gt s_{n}$$ Thus, you need only find the first $n$ such that $s_{n}s_{n+1} \lt 0.01$. Now, if $a_{k+1} \lt 0.01$ then $s_{n}  s_{n+1} = a_{k+1} \lt 0.01$, so the job is done. Last edited by v8archie; May 7th, 2016 at 09:18 PM. 

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approximation, finding, series 
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