Mu Sequences and Series Topic Test Solutions FAMAT State Convention 2018

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1 . E The first term of the sum has a division by 0 and therefore the entire expression is undefined.. C We can pair off the elements in the sum, ( (k + ) + (k + )) = 3. D We can once again pair off the elements to get, 4. A We can rewrite the sum as, = 49 (k + ) + (k + ) = 49 ( k + k + 3 ) then realize that this telescopes such that it equals, k 3 k = C We can multiply by 00 to get that if the number is x, then, 00x =. and then subtract the repeating decimal x to get 99x =. which we can then divide over to get x =. 99 = 990. = 009. (3 + 4k) = B By the alternating series test, this series converges. However, the absolute value of it diverges by it comparison to the harmonic sum. (In fact, this behaves in the same way as the alternating harmonic series convergence-wise.) 7. D We can calculate that, and such that the answer is = = A Simply apply the it comparison test with x.005 which is a convergent p-series to establish that this series converges absolutely.

2 9. C All the elements of the sequence are positive, so the series either converges absolutely or diverges. We can see that it telescopes such that, n ( k + k) = n + which increases without bound, such that the series diverges. 0. C Statement is true simply by separating out the sums. Statement II is not necessarily true because it ignores all the cross terms (so if a k, b k > 0 for all k, it will not be true), and III is true by the n th term test.. A By the AM-GM inequality, we know that, ak k (a k + k ). The sum made up of the RHS converges because we know that a k converges and k is a convergent p-series. Thus, by the comparison test, the series must converge.. D The k+ k part will go to, which means that for large enough k, a k essentially alternates between plus and minus. Thus, the it does not exist. 3. C Using the equation x = n + x, it can be seen that all such n are of the form x x for integer x. Inspection shows that the largest such n < 08 is 980 = B We can recognize this as an arithmetico-geometric series. In particular, we can first split it up, 3 k + k 3 k. The first term is just a geometric series which sums to /. The second term can be written as, 3 k = 3. i= k=i Summing these together we get that the sum evaluates to. 5. C Let n be the number of coins that come up heads if you flip 0 coins. The expected value of n can be calculated as the sum, 0 0 ( 0 k )k. We also know by symmetry of heads and tails that the answer must be 5. Therefore, we know that the sum must evaluate to, 5 0 = 50.

3 6. A In the it, we would need to have that if the ratio is x, then x L n xl n + L n x = x +. This has two roots, but one is negative and makes no sense since the Lucas numbers are monotonically increasing, so the answer must be, B Shifting the indices of the summand, the following result can be obtained. Solving for Φ(x) yields Φ(x) = Φ(x) = L 0 + L x + n= L n x n = + x + (L n + L n )x n = + x + n= n= L n x n+ + n=0 L n x n+ = + x + x(φ(x) ) + x Φ(x) x x x and Φ ( ) = 6 8. A We can take the series representation and write it as a function to get, Then, we can differentiate to get, f(x) = d dx x sin(x). f(x) = sin(x) + x cos(x) and evaluate at x = π to get. 9. C Because + k for all positive integers k, this means that we can bound (k + )a k k a k = 0. This shows that we cannot do any better than 0. Let a = 0 and a k = 0 for all other k. This a k attains our upper bound of 0 making it the maximum possible attainable. 0. D The Taylor series of tan (x) is which we can plug into the it to get, tan (x) = x x3 3 + x5 5 x7 7 + O(x9 ) x O(x ) = 7. 3

4 . B The alternating harmonic series a k = ( ) k k is a counterexample to both I and III. By the n th term-test, k a k = 0 and by it laws a k + = k which means that the series must diverge to infinity.. D If α = 0, then this is simply the harmonic sum and it diverges. For α =, by the integral test using a u-sub, u = ln k, the series diverges. For α >, by the integral test with the same u-sub, the series converges. 3. A The sum can be split through partial fractions to be equal to 8 n ( 0 n 8 n 8 n 0 n+ ). This sum telescopes and terms approach 0 as n 8n+ n= grows large, so plugging in n = gives the sum. This value is 9 4. C For I, consider f(x) = sin(πx) and a k = 0 as a counterexample. II is true because 5. A We can apply the root test by taking, to see that it converges absolutely. a k b k max( a k, b k ) a k + b k. k k/k = 0 6. A For numbers with k digits there are < 0 k/ palindromes if k is even and 0 (k+)/ palindromes with k is odd. The reciprocal of these palindromes are < 0 (k ). The sum of all of the reciprocals of palindromes is < 0 k/. The sum for all positive integer k is a geometric series with ratio <, so the sum absolutely converges. 7. C We can rewrite this as, n n n ( k n ) ln ( k n ). Then, we can recognize this as the definition of the integral We can integrate, 0 x ln x dx. = x 4 + x ln x = D Using Stirling s approximation, we can find that j! e j j j j+0.5 = π and so by the ratio test, the radius of convergence is equal to the reciprocal of this, or 0 π. 4

5 9. B Another way of thinking about 33 ( 99 3k ) is as the sum of the coefficients of f(x) = ( + x) 99 where x is raised to a power that is a multiple of 3. We can find the sum of all coefficients by taking f(), and we can find the sum of all coefficients where x is taken to an even power by taking ((f() + f( )). Similarly, we can find the sum of all the ones that are multiples of 3, by evaluating where ω = e πi/3. f() + f(ω) + f(ω ) A They key to approach this problem is to realize that the terms being summed look like the PDF of a negative-binomial distribution. In particular, because we are taking larger and larger n, we can conveniently replace the negative-binomial distribution with a Normal approximation. In particular, we have the normal approximation N ( n, n ). Then, we are looking for the probability of being further than one standard deviation to the left, for which we can remember the estimate, which means that = 6% of observations will be further than one standard deviation to the left, and thus our series will converge approximately to 6. If you re wondering how good of an estimate this is, evaluating for n = 000 gives 6. and evaluating for n = 0000 gives