Practice for Exam 2 - Math 3410
Intro to Analysis - Spring 2016
1. Determine whether each of the following series is convergent (absolutely or conditionally) or divergent. Indicate
which test(s) you are using, and show all your work.
∞
∞ p
X
X
(a)
ne−n
(b)
( n2 + n − n)
n=0
n=1
∞ p
X
(c)
( n2 + 1 − n)
(e)
n=1
∞
X
n=1
sin n
n2 + 1
(d)
(f)
∞
X
n!
1
·
3
·
5
·
.
.
. · (2n − 1)
n=1
∞
X
cos nπ
n
n=1
(g)
∞
X
log
n=1
n+1
n
2. Determine whether the series below is convergent or not
3 3 3
3
3
− + −
+
+ ...
2 4 8 16 32
If convergent, find its sum.
3. Prove that lim sn = 3 where the sequence (sn ) is defined by
sn =
1
3
5
2n − 1
+
+ 3 + ··· +
,
2 22
2
2n
n = 1, 2, 3, . . . .
(Hint: see Problem 14.13(c)).
4. Show that
∞
X
log n
diverges when 0 < p < 1 and converges when p > 1. What happens when p = 1?
np
n=1
5. Prove that the monotonicity
assumption in the Alternating Series Test is essential. That is, show that there exists
P
an alternating series (−1)n an , for which its sequence of positive terms (an ) satisfies lim an = 0, but which is a
divergent series.
P
P
an converges and
bn converges absolutely, then
an bn converges absolutely.
P
P
P
(b) Construct two convergent series
an and
bn for which the series
an bn diverges.
P
P√
(c) Suppose
an is a convergent series of positive terms. Show that the series
an /n is also convergent.
6. (a) Show that if
P
P
7. Suppose an alternating series (−1)n+1 an , an > 0 and (an ) monotone, converges to a sum s. Show that the
remainder Rn = s − sn satisfies |Rn | ≤ an+1 where sn is the nth partial sum of the series.
8. Suppose a series
∞
X
an is convergent. Show that for every ε > 0, there exists N such that the remainder
n=1
∞
X
an ≤ ε
n=N +1
9. Define
1
1 1
+ + · · · + − ln n .
2 3
n
Prove that an is a decreasing sequence and that 0 < an < 1, hence the limit exists and is a number between 0 and
1. (Hint: See Problem 16.9)
an = 1 +
10. [18 pts] Determine the radius of convergence and the interval of convergence for each of the power series
(a)
∞
X
n=1
n2 xn
(b)
∞
X
(−1)n 2n
x
n
n=1
(c)
∞
X
(4x − 3)n
n2 2n
n=1