Testing for Convergence or Divergence

1 Testing for Convergence or Divergence of a Series Many of the series you come across will fall into one of several basic types. Recognizing these types will help you decide which tests or strategies will be most useful in finding whether a series is convergent or divergent. If nahas a form that is similar to one of the above, see whether you can use the comparison test: Geometric Series = 11nnar convergent if 1<r divergent if 1 r p-Series =11npn convergent if 1>p divergent if 1 p Example: =+121nnn Pick 21nbn=(p-series) 2211nnnan +=, and =121nn converges, so by (i), =+121nnn converges. Some series will obviously not converge recognizing these can save you a lot of time and guesswork. Test for Divergence If 0lim nna, then =1nnais divergent.

2 Example: =+ 1221nnnn nnnannn+ = 221limlim011111lim2 =+ = nnn so =+ 1221nnnn is divergent. Limit Comparison Test (Warning! This only works ifnaand nb are always positive.) If 0lim>= cbannn (and c is finite), then naand nb either both converge or both diverge. Example: = 1121nn Picknnb21=(geometric) 12121limlimnnnnnnba = 012111lim>= = nn =121nn converges, so = 1121nn converges. Consider a series nbso that the ratio nnba cancels the dominant terms in the numerator and denominator of na, as in the example to the left. If you know whether nbconverges or not, try using the limit comparison test. Comparison Test (Warning! This only works ifnaand nb are always positive.) (i) If nnba for all n, and nbis convergent, then nais convergent.

3 (ii) If nnba for all n, and nbis divergent, then nais divergent. If nacan be written as a function with a nice integral, the integral test may prove useful: Integral Test If nanf=)(for all n and )(xfis continuous, positive, and decreasing on [) ,1, then: If 1)(dxxf converges, then =1nnaconverges. If 1)(dxxf is divergent, then =1nnais divergent. Example: =+1211nn 11)(2+=xxf is continuous, positive, and decreasing on [) ,1. dxxdxxtt +=+ 121211lim11]4tanlimtanlim111 == txttt442 = =, so =+1211nn is convergent. Example: = + 13211)1(nnnn (i)211nnbn+=, so 1)1(11121+>+++=+nnnbn nbnn112=+ , so nnbb111 +, so nnbb +1 (ii)011lim1lim232=+=+ nnnnnn So = + 13211)1(nnnn is convergent. Alternating Series Test If (i)nnbb +1for all n and (ii)0lim= nnb, then = 11)1(nnnbis convergent.]

4 The following 2 tests prove Convergence , but also prove the stronger fact that na converges (absolute Convergence ). Ratio Test If 1lim1<+ nnnaa, then nais absolutely convergent. If 1lim1>+ nnnaa or =+ nnnaa1lim, then na is divergent. If 1lim1=+ nnnaa, use another test. Root Test If 1lim< nnna, then nais absolutely convergent. If 1lim> nnna or =+ nnnaa1lim, then na is divergent. If 1lim= nnna, use another test. Example: = 1!nnne !)!1(limlim11neneaannnnnn + += =+= 1lim1nen, so = 1!nnne is divergent. Example: =+1313nnnn 313133lim3limnnnnnnnn + = == nnn13lim271 So =+1313nnnn is divergent. When na contains factorials and/or powers of constants, as in the above example, the ratio test is often useful.

5 Testing for Convergence or Divergence of a Series (continued)