A common mistake was to state that if Ratio test is inconclusive,
convergence is not absolute. This is not right - Ratio test isn't
necessary to detect absolute convergence. Absolute convergence
simply means that the series converges when you replace its terms
with their absolute values. You can then study the series of absolute
values by any method. The Ratio test considers absolute values
(because of the way it's set up), so a convergence outcome of the
Ratio test automatically implies absolute convergence. But the
failure of the Ratio test
doesn't imply anything.
(a) Answer: the series is not absolutely convergent because the series of absolute values is a p-series with p=2/3. However, the series is conditionally convergent by the Alternating Series Test. When applying the Alternating Series Test, you need to explain why the sequence is decreasing (this must involve ALL terms, not just saying that b2≤ b1. (See also comment to Question 2).
(b) The series is convergent by the Comparison Test. Since the terms are all positive, convergence also implies absolute convergence. Few people stated that the terms were positive; some omitted checking convergence of the series they used for comparison.
(c) Many students made mistakes simplifying the expression (esp. factorials) in the Ratio test.
(a) The sequence converges to 1/2. For full credit, you had to include the algebra justifying the limit calculation. Unfortunately, a number of students also confused convergence of sequence with convergence of series.
(b) The sequence is bounded below by 0 (because the terms are positive) and above by 1 (because the numerator is always smaller than the denominator, thus the quotient is less than 1). There is no need to find best possible upper/lower bounds.
There was a lot of confusion on this question -- many students attempted to study the sequence for increasing/decreasing instead of upper/lower bounds. Many incorrectly claimed that the sequence decreases (checking that the first term is greater than the second), and then concluded that the sequence is decreasing from 2/3 (the first term) to 1/2 (the limit), with 2/3 and 1/2 the upper (resp. lower) bounds. In fact, the sequence does the following: it decreases for the first three terms (2/3, 1/2, 10/21 - note that 10/21 is less 1/2), then it increases for the rest of the sequence, to the limit 1/2. To detect this, you'd have to study the entire sequence, not just a couple of first terms -- but this was not required in the question.
(c) Many people computed a1 and a2 instead of s1 and s2. Remember that sn is a partial SUM - thus s2 = a1+a2. Misinterpreting the question this way cost you 3 points. (By constrast, if you know what partial sums are but make a mistake in calculations, you only lose 1 pt). Partial sums are defined for any series and generally have nothing to do with particular formulas valid for geometric series.
(d) The series diverges by Divergence test.
(e) The sequence of partial sums is bounded below by 0, increases, and is not bounded above. This is because partial sums are sums of positive terms of the series - you keep adding more and more terms, so the sequence increases. The series diverges, this sn goes to infinity, and so cannot be bounded above. (Even with correct answers, you had to include explanation to receive full credit.)
(f) Same as (d). Alternating test is irrelevant, and in fact doesn't apply since the sequence is NOT decreasing. The failure of Alternating test doesn't imply any conclusion - you really need to say that the terms of the series do not go to 0, which implies divergence.
For questions about the radius/interval of convergence for power series, use Ratio Test. (A number of people tried to argue that the given series was geometric, but it wasn't.) When the Ratio test gives lim=1, the test is "inconclusive", which means just that -- you can't make any conclusions. A few students incorrectly claimed that lim=1 implies divergence. Note that the Ratio Test NEVER works on endpoints of the interval of convergence (those need to be checked by other tests).
After you found the interval of convergence, there is no need to test points that are inside/outside the interval. However, endpoints need to be tested separately.
(a) This series is divergent. Many students represented it as a sum of two geometric series, one divergent, one convergent. However, for full credit you need to justify why convergent+divergent=divergent series. An easier solution is to apply the Divergence test. Indeed, the first summand in each term oscillates between large positive and large negative numbers, and the second approaches 0 as n grows large.
(b) Convergent series. Represent as a sum of two convergent geometric series to justify convergence and find the sum.
(c) This is also a geometric series, but one that depends on x. You can use the geometric series argument to find where the series converges. Alternatively, you can use the Ratio Test, but for full credit (on "where converges" part) you need to check endpoints separately. The Ratio Test doesn't help to find the sum (for this part, you have to use geometric series).