Mathematical Methods for Physics and Engineering - Matematica.NET

SERIES AND LIMITS◮Given that the series ∑ ∞n=11/n diverges, determine whether the following series converges:∞∑ 4n 2 − n − 3n 3 +2n . (4.13)n=1If we set u n =(4n 2 − n − 3)/(n 3 +2n) andv n =1/n then the limit (4.12) becomes[ ] [ ](4n 2 − n − 3)/(n 3 +2n) 4n 3 − n 2 − 3nρ = lim= lim=4.n→∞ 1/nn→∞ n 3 +2nSince ρ is finite but non-zero and ∑ v n diverges, from (i) above ∑ u n must also diverge. ◭Integral testThe integral test is an extremely powerful means of investigating the convergenceof a series ∑ u n . Suppose that there exists a function f(x) which monotonicallydecreases for x greater than some fixed value x 0 and for which f(n) =u n ,i.e.thevalue of the function at integer values of x is equal to the corresponding termin the series under investigation. Then it can be shown that, if the limit of theintegral∫ Nlim f(x) dxN→∞exists, the series ∑ u n is convergent. Otherwise the series diverges. Note that theintegral defined here has no lower limit; the test is sometimes stated with a lowerlimit, equal to unity, for the integral, but this can lead to unnecessary difficulties.◮Determine whether the following series converges:∞∑ 1(n − 3/2) =4+4+ 4 2 9 + 425 + ··· .n=1Let us consider the function f(x) =(x − 3/2) −2 . Clearly f(n) =u n and f(x) monotonicallydecreases for x>3/2. Applying the integral test, we consider∫ N( )1−1limdx = lim=0.N→∞ (x − 3/2)2 N→∞ N − 3/2Since the limit exists the series converges. Note, however, that if we had included a lowerlimit, equal to unity, in the integral then we would have run into problems, since theintegrand diverges at x =3/2. ◭The integral test is also useful for examining the convergence of the Riemannzeta series. This is a special series that occurs regularly and is of the form∞∑ 1n p .n=1It converges for p>1 and diverges if p ≤ 1. These convergence criteria may bederived as follows.128