Mean = 101.4
Median = 98.5
Standard deviation = 23.1
Problem 1
(a) x^n/n! tends to 0, so n!/x^n tends to infinity. In this case, x=10^6. The sequence diverges.
Some said x^n grows faster than n!, which was a 9 point deduction.
(b) Taking logs gives 2ln n/n which goes to 0, so the limit is 1.
I accepted as given n^(1/n)-->1, so (n^(1/n))^2-->1. Those that stopped with the limit of the logs=0 got a 3 point deduction.
(c) Multiply and divide by (n+sqrt(n^2-n)). The numerator is now n and the denominator is n+sqrt(n^2+n), which is asymptotically equivalent to 2n. This ratio tends to 1/2 as n goes to infinity.
On each part, I took off 7 points for not giving an adequate reason.
Problem 2
(a) The root test is the obvious choice here. n^{1/n} tends to 1 (see part b of Problem 1), so the n-th root tends to 0<1, so the series converges. Since the terms are all non-negative, it converges absolutely.
(b) The numerator behaves as n^{1/3} and the denominator as n^{3/2}. So the series converges by comparison to the p-series, with p=7/6. Since all terms are positive, it converges absolutely.
(c) Write (ln n)^{ln n} as n^{ln ln n}. (To see this identity, take logs of both sides.) Then compare with a p-series for any p>1, since ln ln n is eventually bigger than any fixed p.
More simply ln(ln n) > 2 for n large enough, so ln n ln(ln n) > 2ln n, or (ln n)^{ln n} > n^2, and absolute convergence follows by comparison with 1/n^2.
Again, I took off 7 points for not giving an adequate reason
Problem 3
(a) The ratio test gives radius of radius of convergence of 2. The series diverges at the endpoints, giving an interval of (-2, 2).
(b) Again use the ratio test. When the factorials are cancelled correctly, the ratio a_{n+1}/a_{n}=((n+1)^{2n+2})/(n^{2n}(2n+2)(2n+1)). More simplifying gives ((n+1)/2(2n+1)) ((1+1/n)^n)^2. The first factor tends to 1/4 as n grows large. The second tends to e^2. So the radius is e^2/4. Using Stirling's approximation to n!, at the endpoints, the terms are asymptotically equivalent to 2sqrt(pi*n), so the interval is (-e^2/4, e^2/4).
I took off 5 points for missing the e^2, a very common mistake.
Problem 4
(a) The coefficient of x^{n+3} in the first summand is (n+2 choose 2)2^n. The coefficient of x^n in the second summand is (n+1)(-1)^n. Shift the summation index of the first sum by three and add to get a_n = (n-1 choose 2)2^{n-3} + (-1)^{n}(n+1) (for n at least 3) and a_0=1, a_1=-2 and a_2=3.
(b) The radius for the first sum is 1/2 and is 1 for the second sum. So the radius is 1/2. Since the binomial series diverges at the endpoints, the interval is (-1/2, 1/2).
(c) a_6 = 87.
I awarded 5 points each for parts (b) and (c). Part (b) was independent of part (a). It was not necessary to know a_n to find the interval of convergence. It was also possible to get full credit on part (c) without the correct solution to part (a).
I gave 10 points for just writing down correctly the two sums. There was a great deal of confusion about shifting summation indices and about binomial coefficients.
Problem 5
(a) Write sin^2 x = (1-cos 2x)/2. So the series is 1/2 - (1/2) sum_{k=0 to infinity} (-1)^k 2^{2k} x^{2k}/(2k)!. Simplifying somewhat, (1/2)sum_{k=1 to infinity} (-1)^{k-1} 2^{2k} x^{2k}/(2k)!.
It is also possible to square the series for sin x, giving a double sum as the series for sin^2.
However, many confused squaring the series with squaring the terms of the series.
(b) d/dx (tan x) = sec^2 x; d/dx (sec^2 x) = 2sec^2 x tan x; d/dx (2sec^2 x tan x) = 4sec^2 x tan^2 x + 2sec^4 x. At x = 0, these are 1, 0, and 2 respectively. Since tan x itself is 0 when x = 0, the Taylor polynomial is x + 2x^3/(3!) = x + x^3/3.
Alternatively, perform long division and divide the series for cos x into the series for sin x. That is, divide 1 - x^2/2 +... into x - x^3/6 + .... These are the only terms required to do the computation up to the third degree.
I deducted 5 points for giving only the x^3 term of the polynomial. (A third degree Taylor polynomial has the 0, 1, 2 and 3 degree terms in it - cf. top of page 501.)
I also deducted 5 points for omitting the 3!
Return to my home page.