Math 25 Solutions to practice problems

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1 Math 5: Advaced Calculus UC Davis, Sprig 0 Math 5 Solutios to practice problems Questio For = 0,,, 3,... ad 0 k defie umbers C k C k =! k!( k)! (for k = 0 ad k = we defie C 0 = C = ). by = ( )... ( k + ) k! (a) Prove that for all ad k, Proof. C k + C k = = C k = C k + C k ( )( )... ( k + ) ( )( )... ( k) + (k )! k! k + ( k) ( )... ( k + ) k (k )! = k ( )... ( k + ) (k )! = C k (b) Prove by iductio o that for all, Ck = C0 + C + C C = k=0 Proof. For = we have C 0 +C = + =, so the claim is true. Let be give, ad assume the claim is true for that value of. The, by usig the result of (a) above, we see that + C + k = + k=0 = + k= C + k + = + Cj + j=0 Ck = + k= ( ) C k + Ck k= ( ) Cj + j=0 = + ( ) + ( ) = = +, ( ) Ck k=0

2 where we used the iductive hypothesis i the trasitio from the secod row to the third. This shows that the claim is true for + ad completes the iductio. Questio (a) Let (a ), (b ) be sequeces of real umbers. For each of the followig idetities, explai what assumptios are eeded to esure that the idetity is valid: i. lim (a + b ) = lim a + lim b ii. lim (a b ) = lim a lim b a iii. lim = lim a b lim b Solutio. By a theorem we leared, i. ad ii. are valid uder the assumptio that the sequeces (a ) ad (b ) are coverget. For iii., oe eeds the additioal assumptio that the limit of (b ) is ot zero. (b) Fid the limit L of the sequece give by a = ( + si()) Prove rigorously that L = lim a, by appealig either to the defiitio of the limit or to kow results about limits. Solutio. First, we brig a to a form more suitable for computig the limit, by otig that a = ( ) 4. + si() Now we apply the theorem referred to i part (a) above, with the sta-

3 dard limits lim lim lim lim 5 = 5, 3 = 0, 0 = 0, 4 lim =, si() = 0. The last limit is a result of a applicatio of the squeeze theorem, sice we have the iequalities si() Puttig these results together, we coclude that (a ) is coverget, ad its limit is equal to lim a = ( + 0)( + 0) = 5 4. Questio 3 (a) State the squeeze theorem. Solutio. See Theorem.0 i Sectio.8 i the textbook. (b) Deote a =. Prove that lim a =. You may use the fact that the iequality ( + x) ( ) x holds for all ad x > 0. Solutio. See Example.33 i Sectio.0 i the textbook. 3

4 Questio 4 (a) Prove that the harmoic series diverges. (b) Let p be a positive real umber. Prove that the series if ad oly if p >. p coverges (c) Prove that the series log = diverges (here, log = l()/ l() deotes the base- logarithm of ). Solutio. For parts (a) ad (b) see Sectio 3.4. i the textbook. For part (c), see the solutio to problem 5 i homework assigmet #9. Questio 5 (a) Evaluate the ifiite series Solutio. Recall the formula for the sum of a geometric series =0 x = x, which is valid for all < x <. The above series is a geometric series with x =, so its sum is (b) Evaluate the ifiite series ( ) = 3/ = Solutio. This is a geometric series with x = /3, so by the formula above its sum is /( /3) = 3/. 4

5 (c) Evaluate the ifiite series Solutio. Note that 4 = = ( )( + ) = So, this is a telescopig sum: 4 = = ( ). + ( ) + ( ) = (Note: to properly prove covergece oe eeds to do this computatio more carefully for the partial sums first; but, here the questio oly asked to fid the value of the series). Questio 6 Give a example of: (a) A diverget series (b) A absolutely coverget series (c) A coditioally coverget series (d) A series of the form series a ad ( ) (a + b ) that is coverget but such that both b are diverget. Solutio. Take a =, b = 5

6 (e) A series of the form ad (a b ) that is coverget but such that b are diverget. Solutio. Take a = b =. (f) A diverget series with bouded partial sums (g) A series ( ) a that is diverget such that lim ( a ) = 0 (i.e., loosely speakig, the sequece beig summed coverges to 0 faster tha /). Solutio. The series =, which appears i questio 4(c) above. log Questio 7 (a) Prove that lim = 0. You may use the fact that the iequality ( + x) ( ) x holds for all ad x > 0. Solutio. Takig x = i the iequality metioed i the questio gives ( ), so, for, 0, ad the claim follows from the squeeze theorem. (b) Prove that coverges. Solutio. Sice lim = 0 as show above, takig ε = defiitio of the limit, we get that there is some N such that 0 for all N. Therefore, for N, = ( ) 6 ( ) = a i the

7 Sice also 0, so /( ) > 0, it follows that =N coverges, by the compariso test. This is differet tha the series give i the questio i that the summatio starts at = N istead of =, but as we kow, the covergece of a series oly depeds o its tail i.e., omittig or addig oly a fiite umber of iitial terms does ot affect the covergece, ad therefore the origial series also coverges. Questio 8 The goal of this problem is to compute the value of the ifiite sum S = = (ad i particular to show that it coverges, which stregthes the result of questio 7(a) above). (a) Defie a ew sequece (x ) whose terms are give by (x ) = {,,,,,,,,,,,,,,,...} Explai why if x coverges, the also coverges ad the values of the two series are the same. Solutio. It is easy to see that the sequece of partial sums of is a subsequece of the sequece of partial sums of x. More precisely, if we deote the we have s = k= k k, t = s = t, s = t 3, s 3 = t 6,. s = t ( )/. x k, Therefore if (t ) coverges, the (s ), beig a subsequece, also coverges to the same limit. 7 k=

8 (b) Rearrage the terms of x by writig the sum as ( ) + ( ) + ( ) +... (This part is more of a thikig questio, requirig you to thik about why this rearragemet makes sese ad why it is valid to perform it). (c) Evaluate each of the iteral sums i the above rearragemet, ad the sum of their values, to coclude that S =. Solutio. This was explaied i class i the last lecture. 8