(1) (a) Let
(b) Let
(c) Let
(2) Suppose
Sample midterm 1
an = n − n. What is a5 ?
br = 2ar . What is b2 ?
ck = ak − ak+1 . Find a formula for ck .
that lim an = 4 and lim bn = 2. Determine
2
n→∞
n→∞
(a) lim an /bn
n→∞
(b) lim sin(πb2n )
n→∞
(c) lim an2
n→∞
√
(3) (a) Find x > 0 such that x =
√ 3 + 2x.
(b) Let a0 = 0 and an+1 = 3 + 2an . Assume that lim an exists. Find
n→∞
lim an .
n→∞
(4) Let an = (2n + 5n )1/n . Calculate lim an using squeeze theorem. (Note
n→∞
that 5n < 2n + 5n < 2 · 5n .)
(5) Using the geometric summation formula, find
∞
X
2
.
n−1
3
n=1
∞
X
1
is convergent and less than 1.
n
n2
n=1
Z ∞
3
(7) (a) Calculate
x2 e−x dx.
(6) Explain why
1
∞
X
n2
converges.
(b) Show that
e n3
n=1
(c*) How good is the 10’th partial sum as an approximate for the infinite
series above?
(8) True or False
(a) If lim an = 10 then there is an integer N such that 9.9 < aN < 10.1.
n→∞
(b) There is a sequence an so that a1 = a2 = a3 = · · · = a10 = 1, but
lim an = 0.
n→∞
∞
X
(c) If 0 < lim an < 1, then 0 <
n→∞
an < 1.
n=0
(d) If 0 < an for all n, then lim an > 0.
n→∞
(e) Let an and bn be two positive sequences and assume that
verges. Furthermore, assume that lim
n→∞
converges.
1
an
= 2. Then
bn
∞
X
n=1
∞
X
n=1
bn conan also
2
(9) Which of these series converge, and which ones diverge. State which test
you’re using (something like “comparison test with . . . ” is sufficient)
∞
X
(a)
2n
(b)
(c)
(d)
(e)
n=1
∞
X
1
n(n
+ 1)
n=1
∞
X
1
n=1
∞
X
n log(n)2
n
2−1
n
n=2
∞
X
n
n=2
2n
3
Solutions
2
a5 = 5 − 5 = 20
b2 = 2a2 = 2(22 − 2) = 4
ck = ak − ak+1 = (k 2 − k) − ((k + 1)2 − (k + 1)) = −2k
limn→∞ an /bn = (limn→∞ an )/(limn→∞ bn ) = 4/2 = 2
Since sin is continuous everywhere, limn→∞ sin(πb2n ) = sin(limn→∞ πb2n ) =
sin(4π) = 0
(c) Since n2 → ∞ as n → ∞, we get limn→∞ an2 = limn→∞ an = 4
(3) (a)
√
x = 3 + 2x
(1) (a)
(b)
(c)
(2) (a)
(b)
x2 = 3 + 2x
x2 − 2x − 3 = 0
(x − 1)2 − 4 = 0
x = 3, −1
Since we know x > 0 we get x = 3 is the only solutions to the equation
given.
(b) Let x = limn→∞ an . First note that an > 0, so x > 0. We know that
x = lim an
n→∞
= lim an+1
n→∞
√
= lim 3 + 2an
n→∞
q
= 3 + 2 lim
n→∞
√
= 3 + 2x
Therefore x = 3
(4) Note that
5n < 2n + 5n < 5n + 5n = 2 · 5n
⇒ 5 < (2n + 5n )1/n < 21/n 5.
Now, note that limn→∞ 21/n = 2limn→∞ 1/n = 1, and therefore limn→∞ 5 =
limn→∞ 21/n 5 = 5. Since 5 < an < 21/n 5 for all n, by squeeze theorem we
get that lim
an = 5.
Pn→∞
∞
a
(5) Note that n=1 arn−1 = 1−r
whenever |r| < 1. If we let a = 2 and r = 13
we get
∞
X
n=1
arn−1 =
∞
X
n=1
2
3n−1
=
2
= 3.
1 − 1/3
(6) Note that for n ≥ 1 we have n1 ≤ 1. Therefore 0 < n21n ≤ 21n . Therefore
P∞
by comparison theorem we get that n=1 n21n is convergent. In fact, the
P∞
sequence of partial sums is bounded from above by n=1 21n = 1. Therefore
P∞
1
n=1 n2n ≤ 1.
4
(7) (a) Let u = x3 , then du = 3x2 dx. Furthermore, when x = 1, u = 1, and
when x = ∞ we get u = ∞. Therefore
Z
∞
3
x2 e−x dx =
1
Z
∞
e−u
u
du
3
−e−u ∞
=
|
3 1
−1
=
.
3e
3
(b) Let f (x) = x2 e−x . Note that this is a continuous function. To see
3
3
that f (x) is decreasing, we can compute f 0 (x) = 2xe−x − 3x4 e−x .
3
Note that e−x is always positive, while 2x − 3x4 is negative for all
x > 1, so f 0 (x) < 0 for x > 1, which means f (x) is decreasing.
2 −x3
Since f (x)
is a continuous, and decreasing function, and
R ∞= x e
since the 1 f (x)dx is convergent, we get that the series convergest.
(c) We approximate the error for the 10’th partial sum approximate.
∞
10
∞
X
X
X
n2
n2
n2
−
=
3
3
n
n
e
e
e n3
n=1
n=1
n=11
Z ∞
3
≤
x2 e−x dx
Z10∞
du
=
e−u
3
3
10
−1000
e
=
3
(8) (a)
(b)
(c)
(d)
(e)
(9) (a)
(b)
(c)
(d)
(We can get a very rough estimate for this number without a calculator,
−1000
by noting that e 3 < 2−1000 < (24 )−250 < 10−250 which means that
at least the first 250 digits are correct! In reality, we have at least the
first four hundred digits being correct.)
TRUE: This is by definition of limits.
TRUE: Consider the sequence an = 1 when n ≤ 10 and an = 0 when
n > 10.
P∞
FALSE: If limn→∞ an 6= 0, then n=1 an doesn’t even converge.
FALSE: Consider an = n1 . Then an > 0 for all n, but limn→∞ an = 0.
TRUE: This is by limit convergence theorem.
DIVERGES: The sequence is not converging to 0.
CONVERGES: The sum telescopes, and we can compute it exactly.
DIVERGES: log(1) = 0, so this sum is bad
with. This was
P∞to begin
1
actually due to a typo. I meant to ask n=2 n log(n)
2 , which would
converge by Integral test.
P∞
DIVERGES: limit comparison with n=1 n1 .
5
P∞
(e) CONVERGES: limit comparison test with n=1 √12n (or anything less
than 2 in the denominator would work). We get that
n
2n
n→∞ √1 n
2
lim
n
= lim √ n
n→∞
2
= lim
n→∞
1
√ √ n =0
ln( 2) 2
(The√second to last equality follows from L’Hopital rule, since both n
n
and 2 are going to infinity.) Since the limit goes to infinity, we get
our series converges.