Series Practice Solutions 1) 111 2 2 2 , if both series are convergent12 is a convergent geometric series, 121Show that 2 is convergent.1 2 2 for n 1 (both positive for all n)1 2nn nnnnnnnrnnn is convergent by the comparison test.1So 1 2 is convergent.It is absolutely convergent because the terms are all positive.nnn 2) 1122112211221212121221sin 1211 for all nlim 01 is convergent by the Alternating Series Test1 is a divergent p-series, p = 12So 1 is conditionallynnnnnnnnnnnnnnnn convergent. 3) ln 1Use LCT to compare to (both series have positive terms)2ln2lim lim ln12ln 1So is of higher order than 21 is a divergent p-series, p = 1 (not greatennnnnnnnnnnnnnnr than 1)lnSo is divergent by the LCT.2nn 4) 1lim ln ln31 3lim 1 ln does not exist (and is not equal to 0)31So 1 ln is divergent by the Test for Divergence31nnnnnnnnnn 5) 222212 1 212212Use Ratio Test.2a222222 111lim 2 1So the series is divergent by the Ratio Test.nnnnnnnnnnnnnnnnanannnannnaa 6) 111112!9Use Ratio Test.2!921!992!22!9 2! 922212!1 2! 92221 9lim 1So the series is divergent by the Ratio Test.nnnnnnnnnnnnnnanaannnnnnnnnaa 7) 11 11135 2 15!Use Ratio Test.135 2 15!135 2 1 1135 2 15! 5 !51!13521135215 1!135 2 1 2 15! 135 2 1 5 1!21 1nnnnnnn nnnnnnnannnannan n nnnnnnnn1115121 5( 1)2lim 15So the series is absolutely convergent by the Ratio Test.nnnnnnaa 8) 1131113311 for all n3( 1) 31lim 0311 is convergent by the Alternating Series Test31 is a divergent p-series, p = 1 (not > 1)31So 1 is conditionally convergent.3nnnnnnnnnnnnnn 9) 13131313lim 0 11 is absolutely convergent by the Root Test3nnnnnnnnnnnnanananan 10) 22222222112Use the Root Test.11212121lim 12So the series is absolutely convergent by the Root Test.nnnnnnnnnnnnnnnnnnannannana 11) 4444334444411Prove absolute convergence.1111Use LCT to compare to . (Both series have positive terms.)1 11 1 lim 1 (finite and not equal to 0)1 So the tnnnnnnnnnnnnnn nnnnn344wo series are of the same order.1 is a convergent p-series, p = 2 >1 is convergent by the LCT.1So 1 is absolutely convergent.1nnnnnn 12) ln1ln ln1ln 1Compare to using Comparison Test.ln 1 for n > e (Both series have positive terms.)11 is a divergent p-series, (not greater than 1)2ln is divergennnnnnnnnnnnnnpnnt by the Comparison Test.lnSo 1 is not absolutely convergent.Use Alternating Series Test to prove series is conditionally convergent.ln 1ln Show that 1ln Let ( ) nnnnnnnxfxx1222211 1111ln 1lnln222' So 0 for ln 2, or .ln 1ln Therefore for .1ln lim 0ln So 1 is convergent by the ASTnnxxxxxxxfxxxxxfx x x ennnennnnnn.lnSo 1 is conditionally convergent.nnn 13) 221122111Use Integral Test.ln lnlim ln 11 ln 1 limln 1 limtttttttxxdx dxxxuxdvxdxdu dx vxxxdxxxxxx 212ln 1 ln1 lim 11 0 0 0 1 1 ln is convergent.lnSo is convergent by the Integral Test.
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