Transcription of Math 115 Exam #1 Practice Problems
1 math 115 Exam #1 Practice ProblemsFor each of the following, say whether it converges or diverges and explain n=1n3n5+3 Answer:Notice thatn3n5+ 3<n3n5=1n2for alln. Therefore, since 1n2converges (it s ap-series withp= 2>1), the series n3n5+3alsoconverges by the comparison n=13n4n+4 Answer:Notice that3n4n+ 4<3n4n=(34)nfor alln. Therefore, since (34)nconverges (it s a geometric series withr=34<1), the series 3n4n+4also converges by the comparison n=1n2nAnswer:Using the Root Test:limn n n2n = limn n nn 2n= limn n n2= the limit is less than 1, the Root Test says that the series converges what values ofpdoes the series n=1np2+n3converge?
2 Answer:Doing a limit comparison to 1n3 p, I see thatlimn np2+n31n3 p= limn n32 +n3= , the series converges if and only if the series 1n3 pconverges. This happens when 3 p>1,which is to say whenp<2. So the given series converges whenp< would like to estimate the sum of the series n=11n4+3by using the sum of the first ten terms. Ofcourse, the exact error is the sum of all the terms from the 11th on, , n=111n4+3. Show that thiserror is less than 1/3000 by comparing this with the sum of 1/n4and then by estimating this lattersum using an appropriate :Notice that1n4+ 3<1n4for alln, so n=111n4+ 3< n= turn, the sum on the right is less than 101x4dx= 13x3] 10=13000,so we see that the error is less than 1 the series n=1n!
3 (n+ 1)!(3n)!converge or diverge?Answer:Using the Ratio Test,limn (n+1)!(n+2)!(3n+3)!n!(n+1)!(3n)! = limn (n+ 1)!(n+ 2)!(3n+ 3)! (3n)!n!(n+ 1)!= limn (n+ 1)(n+ 2)(3n+ 3)(3n+ 2)(3n+ 1)= limn n2+ 3n+ 227n3+ 54n2+ 33n+ numerator and denominator byn3yieldslimn 1n+3n2+2n327 +54n+33n2+6n3= 0<1, the Ratio Test says that the series converges the series n=1( 1)ncos(1n)converge absolutely, converge conditionally, or diverge?Answer:Notice thatlimn cos(1n)= limx cos(1x)= cos(limx 1x)= cos(0) = 1since cosine is a continuous function.
4 Therefore, the terms( 1)ncos(1n)are not going to zero, so the Divergence Test says that the series the radius of convergence of the series n=0n3x3nn4+ 1 Answer:Using the Ratio Test,limn (n+1)3x3n+3(n+1)4+1n3x3nn4+1 = limn (n4+ 1)(n+ 1)3x3n3((n+ 1)4+ 1) =|x|3limn (n4+ 1)(n+ 1)3n3((n+ 1)4+ 1)=|x|3limn n7+..n7+..=|x|3which is less than 1 when|x|<1, so the radius of convergence is the sequence defined byan=( 1)n+n( 1)n n. Does this sequence converge and, if it does, to whatlimit?
5 Answer:Dividing numerator and denominator byn, we have thatlimn ( 1)n+n( 1)n n= limn 1n(( 1)n+n)1n(( 1)n n)= limn ( 1)nn+ 1( 1)nn 1=1 1= 1,so the sequence converges to the value of the series n=11 + 2n3n :I can re-write the terms as:1 + 2n3n 1=13n 1+2n3n 1=(13)n 1+ 2(23)n , n=11 + 2n3n 1= n=1(13)n 1+ n=12(23)n the indices of the sums down by one yields n=0(13)n+ n=02(23) are both geometric series, so I can sum them using the formula for geometric series: n=0(13)n+ n=02(23)n=11 13+21 23=32+ 6 = the series n=1n+ 5n n+ 3converge or diverge?
6 Answer:Do a limit comparison to 1 n:limn n+5n n+31 n= limn (n+ 5) nn n+ 3= limn n3/2+ 5n1/2 n3+ numerator and denominator byn3/2yieldslimn 1n3/2(n3/2+ 5n1/2)1n3/2 n3+ 3n2= limn 1 +5n 1n3(n3+ 3n2)= limn 1 +5n 1 +3n= , since 1 n= 1n1/2diverges (it s ap-series withp= 1/2<1), the Limit ComparisonTest says that the given series also the series n=13 + cosnenconverge or diverge?Answer:Notice that|3 + cosn| 4for alln, so 3 + cosnen =|3 + cosn|en 4en= 4(1e)nfor alln. Since1e<1, the series 4(1e)nconverges and so, by the comparison test, 3+cosnen , the series 3+cosnenconverges the series n=0( 1)n1 n2+ 1converge absolutely, converge conditionally, or diverge?
7 Answer:The terms1 n2+1are decreasing and go to zero (you should check this), so the AlternatingSeries Test says that the series see that the series does not converge absolutely, it suffices to show that the series n=0 ( 1)n1 n2+ 1 = n=01 n2+ 1diverges. To see this, do a limit comparison with the divergent series 1n:limn 1 n2+11n= limn n n2+ 1= limn 1nn1n n2+ 1= limn 1 1n2(n2+ 1)= limn 1 1 +1n2= the limit is finite and non-zero, the limit comparison test says that the series 1 n2+ the series n=1( 1)nn!
8 Nconverge absolutely, converge conditionally, or diverge?Answer:Using the Ratio Test,limn ( 1)n+1(n+1)! n+1( 1)nn! n = limn n+ 1 = .Therefore, the Ratio Test says that the series
