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Unformatted text preview: Math 182 Homework 8.3
Problem 12
Determine whether the series is convergent or divergent: n=1 5 4 + n4 n n We will use the integral test: 1 5 4 + 4 x x x dx =
1 5x4 + 4x 2 dx =  3 5 3 1 x + 4x 2 3 x= =
x=1 5 3 1 x + 4x 2 3 x=1 x= 4 17 5 5 12 + = + 0= < 3(+)3 3 3 3 + Therefore, since the improper integral converges, so does the infinite series. = 5 +4  3 Problem 20
Determine whether the series is convergent or divergent: n=1 n=1 an = 1 n3 + 1 n=1 bn Here we will use the limit comparison test with the series that this latter series converges: 1 =
x=1 1 n=1 3 n2 . Now using the integral test we see x 2 dx =  2x 2 3 1 x= x=1 = 2x 2 1 x= =2< For limit comparison we have the following: an = lim = lim n bn n 1 1+n3 1
3 n2 3 n2 n3 = lim = lim = lim 3 n n 1+n 1 + n3 n n3 n3 = +1 n n3 lim n3 = 1=1 +1 This result is nonzero and finite. This means that the convergence or divergence of both series is the same and hence we do have that our original series does converge. Problem 36
Show that if we want to approximate the sum of the series n1.001 so that the error is less than 5 in the ninth n=1 decimal place, then we need to add more than 1011,301 terms. We note that if we add up terms until n = N then the error in using n=1 n1.001 in place of be no more than x= x0.001 x=N x1.001dx =  = 1000x0.001 x= = 1000N 0.001 0.001 x=N N
0.001 N 1.001 n=1 n will Therefore, we need that 1000N 0.001 < 5 109 . Therefore, we have that N > .2 109 . Hence N 0.001 > 1000 1000 0.2 109 103 = 0.2 1012. So we need that N > 0.001 2 1011. This means that N > 2 1011 . This is a huge number, we can get an idea of which power of 10 we have by taking log10 of the result. Therefore, log10 N > log10 (2 1011 )1000 = 1000 log10 (21011 ) = 1000 log10 2+1000 log10 (1011 ) = 1000 log10 2+100011 11, 301 and hence we have that N > 1011,301 as claimed. 1 ...
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This note was uploaded on 04/08/2008 for the course MATH 182 taught by Professor Keppelmann during the Spring '08 term at Nevada.
 Spring '08
 Keppelmann
 Math

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