Math 1105: Calculus II (Math/Sci majors) MWF 11am / 12pm, Campion 235 Written homework 10

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1 Math 05: Calculus II Math/Sci majors) MWF am / pm, Campion 35 Written homewor ,6,68,73,80,8,84), 86,58), Chapter 8 Review60,7) Problem 56, 85 Use a test of your choice to determine whether = ) Solution In order to use the Root Test, we tae the th root of the terms in the sum: ) ) / = ) Taing the it as approaches of these terms is a bit difficult with the in the eponent, so we tae the logarithm: log ) ) log ) log ) ) log The latter is indeterminate as approaches, so we tae the derivatives of the numerator and denominator in order to use l Hôpital s Rule This implies that so that log = ) ) =, ) = e Since e <, the Root Test implies that the series converges Problem 6, 85 Use a test of your choice to determine whether / Solution We will perform the Divergence Test Since the variable appears in the eponent, we tae the logarithm again to compute the it ) log / log / log The latter is indeterminate as approaches, so we tae the derivatives of the numerator and denominator in order to use l Hôpital s Rule This implies that = 0 log = 0, =

2 so that / = In particular, the series diverges by the Divergence Test Problem 68, 85 Use a test of your choice to determine whether converges or 4 diverges Solution Evidently, the pattern that determines the term in the series can be described by letting a = / + ) = / + + ), so that the series is + + In order to apply the Limit Comparison Test, we let b = /, and we compute a b ++ By the Limit Comparison Test, the series = + + a and = = + + = the p-test, the harmonic series =/ diverges, so that the given series diverges b either both converge or both diverge By Problem 73, 85 Find the values of the parameter p > 0 for which the series ) p log con- verges = Solution Following the Eample 4b) on page 646, we eamine the quotient of log p / p by a : log p p a = log p p a As long as p a > 0, the it of the latter will be zero see Theorem 86 on page 64) For any a < p, the Limit Comparison Test would then say that the given series converges as long as the series a = converges, which will happen as long as a > by the p-series test For any p >, we can choose any value for a between p and for instance, + p)/ will do), and the argument above implies that the original series converges If on the other hand p, we can compare log /) p to log / directly Since log / <, we must have ) p log log This can be seen using calculus: When p <, p ) = pp ) p <, so p is concave down Since the function p is zero at = 0 and =, the function is positive for in 0, ) Thus p > when p < and is in 0, ) ) p log This means that log for p Since the harmonic series diverges, the Comparison ) p log Test says that diverges for p =

3 Problem 80, 85 Determine whether the following series converge: a) sin b) = = sin Solution a) We have sin sin 0 + =, where we ve substituted = / which goes to 0 as goes to By the Limit Comparison Test, the series sin and either both converge or both diverge Since the harmonic series diverges, the = = given series diverges Solution b) We have sin sin = By the Limit Comparison Test, the series sin and either both converge or both diverge = = Since the latter series converges by the p-series test, the given series converges sin Problem 8, 85 Use the Ratio Test to determine the values of 0 for which the series converges =! Solution We have In order to apply the Ratio Test we have + + )!! = + + = 0 for any This implies that the series converges for any Problem 84, 85 Use the Ratio Test to determine the values of 0 for which the series converges = Solution We have + + = + This means that + =, so that the series converges as long as < If = the series becomes the harmonic series and so diverges, but if = the series becomes the alternating harmonic series and so converges This means the series converges for in [, ) and diverges for other values of

4 Problem, 86 Determine whether = cos π converges or diverges Solution When is even we have cos π = whereas for odd we have cos π = This means that cos π = ) Since {/ } is decreasing and has it 0, the Alternating Series Test implies that the given series converges Problem 58, 86 Show that the series diverges Which condition of the Alternating Series Test is not satisfied? ) + + = Solution Note that we have + + =, so that the terms ) + / + ) do not approach 0 Since it s not true that the it of the terms is 0, the Divergence Test implies that the series diverges Problem 60, Chapter 8 Review Show that the series = converges provided p > log ) p Solution In order to use the Integral Test, we chec that the function f) = / log p ) is decreasing: Since for all > 0, f is decreasing i f ) = logp + p log p log p This means that the series < 0 = log p and the improper integral d nfty log p either both converge or both diverge We use substitution to compute the latter: d a log p d log a a log p du a u p a ) u p log a p a p log log a p ) = log p a log p log p a log p ) p ) log p In particular, this improper integral converges, so that the original series converges by the Integral Test Problem 7, Chapter 8 Review Let a n = ma{sin, sin,, sin n}, for n =,, 3,, where ma{ } denotes the maimum element of the set Does {a n } converge? If so, mae a conjecture about the it Solution Because a n+ is the maimum of the list of numbers {sin, sin,, sin n, sinn + )} and a n is the maimum of {sin, sin,, sin n}, we must have that a n+ a n In particular, the sequence {a n } is increasing technically, non-decreasing) On the other hand, for each n we have a n = sin for some integer, so that a n for each n Since the sequence {a n } is increasing technically, nondecreasing) and bounded above, the Monotone Convergence Theorem guarantees that the it eists

5 In fact, with more wor it can be shown that the it is 5