MATH1220: Midterm 3 Practice Problems
The following are practice problems for the second exam.
1. Compute the following limits:
x2 − 2x + 1
=0
x→1
sin(πx)
(a) lim
ex − e−x
=1
x→0 2 sin x
(ln x)2
(c) lim
=0
x→∞
2x
x
1
(d) lim 1 +
=e
x→∞
x
(b) lim
2. Evaluate the following integrals:
Z 1
1
(a)
e4x dx= e4
4
−∞
Z ∞
x
dx= ∞
(b)
1
+
x2
5
Z ∞
2π
dx
=
Obtain this answer usinga trig substitution.
(c)
2 + 9)2
(x
81
−∞
Z 3
dx
(d)
dx Answer: Diverges
2 − 2x − 3
x
0
Z 0
dx
(e)
Answer: Diverges
(x
+
3)2
−4
3. Write an explicit formula for the n-th term of the sequence. Then determine whether the
sequence converges or diverges. If it converges, find what number it converges to:
2
(a) a1 = 7, an+1 = an
Answer: an = 7(2/3)n−1
3
(b) −1, 2, 5, 8, 11, . . . Answer: an = −4 + 3n
1 2 3 4 5 6
n−1
(c) 0, , , , , , , . . . Answer: an =
4 6 8 10 12 14
2n
2n3
2
= .
5n3 − 2n + 2 5
n
1
5. Show that the sequence an =
2 − 2 converges using the monotone sequence theon+1
n
n
1
rem. Answer: an is an increasing sequence because
is increasing as is 2 − 2 . Also, an
n
+
1
n
n
1
2 − 2 ≤ 1 · 2 = 2. Hence it converges.
is bounded above by 2 since an =
n+1
n
4. Find the limit of the sequence an =
1
6. Determine the convergence/divergence of the following series:
∞
X
n
Answer: Diverges by the divergence test (the terms don’t go to zero)
n
+
1
n=1
" k #
∞
k
X
1
1
(b)
−3
5
Answer: Converges because it’s a difference of convergent geo2
7
k=1
5
3
metric series. It converges to
−
1 − 1/2 1 − 1/7
∞
X
2
Answer: Converges by limit comparison test.
(c)
(k + 2)k
(a)
(d)
k=1
∞
X
ln(k/(k + 1)) Answer: Diverges because it’s a collapsing series and the partial sums
k=1
don’t converge.
∞
X
3
Answer: Converges by limit comparison test
(e)
2
2k + 1
(f)
(g)
k=1
∞
X
k=1
∞
X
1000k 2
Answer: Diverges by limit comparison test.
1 + k3
k sin(1/k) Answer: Diverges by the divergence test. lim k sin(1/k) = lim
k→∞
k=1
lim
(h)
(i)
(j)
(k)
k=1
∞
X
k=1
∞
X
k=1
∞
X
sin(1/k)
=
1/k
.
x→0 x
∞ √
5
X
3n4
k=1
(l)
sin x
k→∞
n2
+3
Answer: Converges by limit comparison test.
3k + k
Answer: Converges by ratio test.
k!
ln k
Answer: Converges by ratio test.
2k
42n
Answer: Converges by ratio test.
n!
ln 2 ln 3 ln 4 ln 5
+ 3 + 4 + 5 + . . . Answer: Converges by ratio test. Computing the limit is
22
3
4
5
difficult because it involves several tricks.
7. Determine whether each of the following is absolutely convergent, conditionally convergent,
or divergent:
(a)
∞
X
1
(−1)n √ Answer: Conditionally convergent by alternating series test and absolute
n
n=1
ratio test.
2
(b)
(c)
∞
X
(−1)k+1
k=1
∞
X
(−1)n
n=1
k2
Answer: Absolutely convergent by absolute ratio test.
ek
n−1
Answer: Divergent by alternating series test (terms don’t go to zero).
n
3