Download Monotone Sequence - Real Analysis - Solved Exam and more Exams Mathematics in PDF only on Docsity!
Math 312, Intro. to Real Analysis:
Midterm Exam #2 Solutions
Stephen G. Simpson
Friday, March 27, 2009
- True or False (2 points each)
(a) Every monotone sequence of real numbers is convergent. False. (b) Every sequence of real numbers has a lim sup and a lim inf. True. (c) Every sequence of real numbers has a monotone subsequence. True. (d) Every sequence of real numbers has a convergent subsequence. False. (e) If lim inf an = lim sup an = α then lim an = α. True. (f) We can find a sequence of real numbers, (an), such that the sub- sequential limits of (an) are exactly the real numbers in the closed interval [− 1 , 1]. True. An example is the sequence
(g) The series 1 −
- · · · is absolutely convergent. True. (h) If
|an| is convergent, then so is
|an|^2. True. Here is a proof. If
|an| < ∞, then lim |an| = 0, hence in particular we can find an N such that |an| < 1 for all n > N. Then, for all n > N we have |an|^2 < |an|, so
|an|^2 converges by comparison with
|an|. (i) If
an is convergent, then so is
a^2 n. False. An example is the series
which is convergent by the alternating series test, but the series
is divergent.
(j)
∑^ ∞
n=
np^
is convergent for all p ≥ 1.
False. However, it is true for p > 1.
(k)
∑^ ∞
n=
10 n^
True. This is a geometric series. (l) The function
x is uniformly continuous on [0, ∞). True. (m) If a function is uniformly continuous on the interval (a, b] and on the interval [b, c), then it is uniformly continuous on the interval (a, c). True.
- (6 points each) Which of the following series are convergent and/or ab- solutely convergent? Please indicate which tests you are using and show your work.
(a) 1 −
Solution. Convergent by the alternating series test. Not absolutely convergent, because it is known that the harmonic series 1 + 12 + 13 + 1 4 +^ · · ·^ is divergent (it is the^ p-series with^ p^ = 1).
(b)
∑^ ∞
n=
n^2 + n − n)
Solution. lim(
n^2 + n − n) = 1/ 2 6 = 0, so the series is divergent by the nth term test.
(c)
∑^ ∞
n=
n^2 + 1 − n)
Solution. We have
n^2 + 1 − n =
n^2 + 1 + n
3 n , so the given
series diverges by comparison with the divergent series
3 n
(d)
- 01 n^ − 1000 Solution. Converges absolutely, by the ratio test.
(e)
n^2 +
n^2 + 1 Solution. Converges absolutely, by comparison with the convergent series
n^2
(the p-series with p = 2).
(f)
∑^ ∞
n=
log
n + 1 n
Solution. Note that log
n + 1 n
= log(n + 1) − log n. By cancelling terms, the nth partial sum of our series is
(log 2−log 1)+(log 3−log 2)+· · ·+(log(n+1)−log n) = log(n+1).
Since the limit of the nth partial sum is +∞, our series is divergent.
- (8 points) Use algebra plus limit laws to calculate
lim
log
e(2n+11)/n sin((4nπ + 5)/ 16 n)
In performing this calculation, you may take it for granted that functions such as sin x, cos x, ex, log x,
x, etc. are continuous. Please show your work.
