Solved problems: (Power) series 1. Sum up the series (if it converges) 3 k+1 a) 2 2k+5 ; b) 1. k(k + 1).

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1 Power series: Solved problems c phabala 00 3 Solved problems: Power series. Sum up the series if it converges 3 + a +5 ; b +.. Investigate convergence of the series a e ; c ; b 3 +! ; d a, where a 3. Investigate convergence of the series a p for p 0,,; b 3 +! {, odd,, 3 even. + ; c Investigate convergence and absolute convergence of the series Epand the given function f into Taylor series with the given center 0. a f + e, 0 ; c f + 5 +, 0 ; b f + sin π +, 0 ; d f Solutions, 0. a. This loos lie a geometric series: < b This loos lie a telescopic series, we try it: so s N N N + N N+ N+, +. a. This is a typical problem for the limit ratio test: λ lim a+ a lim + /e + /e Since λ e <, the series converges. e Also the limit root test would wor: We cheated a bit, we should chec that lim + lim a lim e lim e e lim +/ + / e e. lim lim / 0 lim ln e l H lim e e / e 0. e e.

2 Power series: Solved problems c phabala 00 Anyway, we have e < and we get convergence of the series. Integral test is also possible, since the function f e is non-increasing chec its derivative on [, : e d e d u u [ e + 3e lim e ] v e [ v e e ] + l H 3 e 0 + [e 0] 5e. e d Integral came up finite, hence also the series converges. e Since a 0, we have a e a, so convergence and absolute convergence coincide. Conclusion: The series converges absolutely. b. For alternating series we have just one test, the AST. For that we need that b 3 +! be non-negative and non-inreasing. This is fortunately true, for we even have a decreasing sequence: b ! ! 3 +3 b < b. So we can use AST, and since b 0, the series converges. What is the type of convergence, is it also absolute? Does 3 +! converge? Because of the factorial we cannot use the integral test, nor the root test, factorials can be handled only by the limit ratio test: λ lim a+ a lim 3 + /+3! 3 /+! lim +! +3! lim We have λ 0 <, so the series 3 +! converges. Therefore the series 3 +! converges absolutely, not conditionally. By the way, an eperienced series solver would guess that the factorial in the denominator would be decisive, so he would go directly to testing absolute convergence via the ratio test, then by a theorem the series itself automatically converges and we don t have to use AST. c. For alternating series we have only one test, namely AST, but for that we need b to be non-increasing. However, f is increasing for 3, so AST is out. We can try for absolute convergence, does converge? Here we can use both the root and ratio test, we start with the former: a [ ] as. We have >, so the series diverges. Thus the series does not converge absolutely. The limit ratio test would yield the same answer: a + a + /+ + / + +/ pro. We still do not now whether the series converges but not absolutely or diverges. Since we have no other tests for alternating series, it will probably diverge, which is something we can tell from the limit of a. And indeed, [for instance using two l Hospital rules], so it is definitely not true that a ± 0. The given series therefore diverges.

3 Power series: Solved problems c phabala 00 d. We have a series with non-negative numbers again, so convergence and absolute convergence coincide and we can use our favorite tests. There are only powers in epressions for a, so both root and ratio tests might wor. However, since even and odd terms are of different types and the ratio test would mi them up, it will be probably easier to try the root test: a {, odd, 3, even. The limit of a does not eist, so the limit root test failed. However, we can try the ordinary root test. If we put q, then q < and we also see that a q for all. By the root test the given series converges also absolutely due to a 0. By the way, how would the ratio test fare? A bit of careful reasoning shows that a + a { , odd, 3 + 3, even. Since 3 3 0, while 3, there is no limit of the epression a + a and the limit ratio test fails. Here we cannot even use the ordinary ratio test, since the terms a + a alternate being greater than and less that, so none of the two variants is true. Ratio failed in any form. 3 a. First we rearrange the series in order to see its center and have it in a nice form: p p [ ] p. The center is 0. Now it s time for absolute convergence, so we need to investigate the series p. Since we have only powers there, the limit root test is probably p best. a [ ] p. p As we can see, so far the value of the parameter p did not play any role. In order for the series with absolute value to converge we need <, which happens eactly if <. Thus we get the radius of convergence r. How about endpoints? They are ±. What if we substitute 5? We get, so we see p that p begins to matter. It is best to do each case separately. Case p 0, the series. We have r, we try endpoints. 5 :, the series diverges. 3 :, the series diverges. Conclusion: The series converges absolutely and converges on 3, 5. Case p, the series. We have r, we try endpoints. 5 :, the series diverges harmonic series. 3 :, the series converges this one we now, it is handled via AST. Conclusion: The series converges absolutely on 3, [ 5, converges on 3, 5. Case p, the series. We have r, we try endpoints. 5 :, the series converges we now this one, too, and if we forget, we use integral test. 3 :, the series converges via AST. So the series converges for both endpoints, hence it converges absolutely at ±. Conclusion: The series absolutely converges and converges on [ 3, ] 5. 3

4 Power series: Solved problems c phabala 00 3 b. First we rearrange the series in order to see its center and have it in a nice form: !! [ + ] 3 3! + 3 6!. The center is 0. Now it s time for absolute convergence, so we need to investigate the series 3 6! 3 6! +. Factorial clearly indicates that we should use the ratio test, we try the limit version: a + a 3 6+ / +! /! ! ! λ. In order for the series with absolute value to converge we need λ 0 <, which is always true. Thus we get the radius of convergence r and the given series converges absolutely and hence also converges on IR. 3 c. First we rearrange the series in order to see its center and have it in a nice form: The center is 0 0. Now it s time for absolute convergence, so we need to investigate the series. Since we have only powers, probably the limit root test will be best a 3 3 {, 0, 3 0, 0. In order for the series with absolute value to converge we need lim a <, which is true only for 0, that is, 0. Thus we get the radius of convergence r 0. Conclusion: The series converges absolutely and converges on the set {0}.. First we rearrange the series in order to see its center and have it in a nice form: [ ] + 8. The center is 0. Now it s time for absolute convergence, so we need to investigate the series + +. Here both ratio and root seem possible, but the 8 8 complicated root will probably be better in a ratio: a + a + + / / / + / + / λ. In order for the series with absolute value to converge we need λ 8 <, which is true for < 8. Thus we get the radius of convergence r 8. Endpoints are ± 8. 0: +, the series diverges, since + and so c + 0 is not true. 6: +, the series diverges. Conclusion: The series converges absolutely and converges on 6,0.

5 Power series: Solved problems c phabala 00 5 a. We need to get : + e + 3e +3 e 3 e + 3e 3 e for y IR IR, r e 3 [ ] + 3e 3 [ ]!! 0 0 e 3 + 3e 3 + +!! 0 0 e 3! 3e 3 +! 0 e 3 3e 3 +!! 0 e 3 [ 3e e 3 + ]! 0!! [3e 3 e 3 + ] + ; IR.! 5 b. We need to get + : + sin π + + sin π + sin π + π + cos π sin α π for y π [ π 0 + ]! π! [ π 0 0 0! cosα, α π + + IR IR, r + π! + +] π! + + ; IR. 5 c. We need to get and get rid of in the numerator [ for y < <, r ] ;,3. 5

6 Power series: Solved problems c phabala 00 5 d. We need to get, but first we need to get rid of a power using a tric with derivative. By guessing or using d : [ ] [ ] [ ] [ ] + [ ] for y < <, r [ [ ] ] [ + ] ;,3. 0 6