Sample midterm 2
(1) The following sequences are convergent. Find their limits.
(a) an = √nn2 +1 .
(b) an = n1/n .
√
(c) a1 = 1, and an+1 = 2 + an for n > 1.
(2) Which of these series converge and why.
∞
X
n
(a)
log
n
+
1
n=1
∞
X
n
(b)
3n
n=1
∞
X
1
(c)
n(log
n + 1)
n=1
∞
X
1
n
(d)
(−1) cos
n
n=1
(3) Calculate the radius of convergence of each of the following power series.
∞
X
nxn
(a)
(n + 1)!
n=0
∞
X
xn
(b)
3n + 2n
n=0
∞
X nxn
(c)
2n
n=0
R dx
−1
(4) (a) Recall that 1+x
x + C. Prove that
2 = tan
tan−1 x = x −
x5
x7
x3
+
−
+ ···
3
5
7
(b) Evaluate
1 1 1
+ − + ···
3 5 7
R1
(5) (a) Let J = 0 sin(x2 )dx. Express J as an infinite series.
(b) Determine J to within an error less than 0.1.
(6) Let f (x) = (sin(x), cos(x), x).
(a) Find Rf 0 (x).
x
(b) Find 0 f (t)dt.
(c) Calculate ||f 0 (t)||.
1−
1
2
(1) (a) lim √
n
= lim
n2
+1
limit is 1.
(b) First we calculate
n→∞
n→∞
n
1
p
= lim p
= 1. So the
2
n→∞
n 1 + 1/n
1 + 1/n2
log(n)
.
n
Since both numerator and denominator is going to infinity, we can use
L’Hopital rule. Therefore
lim log(an ) = lim
n→∞
lim
n→∞
n→∞
1/n
log(n)
= lim
= 0.
n→∞
n
1
Therefore
lim log(an )
lim an = lim elog(an ) = en→∞
= e0 = 1.
n→∞
n→∞
(c) Let L = lim an (the question states that they are convergent, so we
n→∞
know such an L exists). Note that
q
√
√
lim an = lim an+1 = lim 2 + an = 2 + lim an = 2 + L.
n→∞
n→∞
n→∞
n→∞
√
2
Therefore L = 2 + L. Therefore
√ L = 2 + L, and hence L = 2 or
L = −1. However note that L = 2 + L > 0, therefore the limit is 2.
n
(2) (a) Note that log n+1
= log n − log(n + 1). Therefore
∞
X
log
n=1
∞
X
n
=
log n − log(n + 1)
n + 1 n=1
which is a telescoping sum. Therefore
SN =
N
X
n=1
log
n
= log 1 − log(N + 1) = log(N + 1).
n+1
Since Sn = lim log(n + 1) diverges, we get that this sum diverges.
n→∞
Remark 1: Initially I made a mistak and claimed this sum converges,
since I didn’t check for the lim log(n + 1) going to zero. Thankfully
n→∞
Reed found the mistake.
Remark 2: Note that n/(n + 1) < 1, therefore log(n/(n + 1)) < 0,
therefore you need to be a bit careful if you try to use comparison
∞
X
n
theorem. However, we can consider
− log
which is positive,
n+1
n=1
∞
X
1
and limit comparison with
does work in this case.
n
n=1
(b) We can use Ratio test. Note that
(n + 1)/3n+1
(n + 1)3n
n+1
1
=
lim
= lim
= .
n
n+1
n→∞
n→∞ n3
n→∞ 3n
n/3
3
lim
Since this limit is less than 1, by Ratio test we get that this series
converges.
3
1
(c) We use Integral test. Let f (x) = x(log(x)+1)
. Note that f is positive
and decreasing, and an = f (n). Therefore by integral test we get that
∞
X
R∞
1
1
dx converges.
converges if and only if 1 x(log(x)+1)
n(log(n)
+
1)
n=1
Let u = log(x), then du = dx
x . When x = 1 we get u = 0 and when x
goes to ∞ we get that u goes to infinity. Therefore
Z ∞
Z ∞
1
du
dx =
= lim log u + 1,
x(log(x) + 1)
u + 1 u→∞
1
0
which diverges. Therefore the series diverges.
(d) Note that lim cos(1/n) = cos(0) = 1. Therefore lim (−1)n cos(1/n)
n→∞
n→∞
does not exists. Therefore the series diverges
(3) (a) We can use Ratio test. Note that
(n + 1)x (n + 1)(n + 1)!xn+1 (n + 1)xn+1 /(n + 2)! = |x| lim (1 + 1/n) = 0.
= lim = lim lim n
n
n→∞ n(n + 2) n→∞
n→∞
n→∞
nx /(n + 1)!
n(n + 2)!x
n+2
Since 0 < 1, we get that the series diverges for all values of x. Therefore
the radius of convergence is infinity.
(b) We can use the power test. Note that
xn 1/n
|x|
|x|
= lim
=
.
lim
n
n→∞ (3n + 2n )1/n
n→∞ 3n + 2n lim (3 + 2n )1/n
n→∞
n
n 1/n
We need to find lim (3 + 2 )
n→∞
. Note that
3n < 3n + 2n < 2 · 3n .
Therefore
3 < (3n + 2n )1/n < 21/n · 3.
Since lim 21/n 3 = 3 and lim 3 = 3, we get that lim (3n +2n )1/n = 3
n→∞
n→∞
n→∞
by squeeze theorem. Therefore
xn 1/n
|x|
=
,
lim
n→∞ 3n + 2n 3
and by power test we get that this seires converges when
|x| < 3. Therefore the radius of convergence is 3.
(c) We can use ratio test Note that
|x|
3
< 1, or
(n + 1)xn+1 /2n+1
(n + 1)xn+1 2n
(n + 1)x
|x|
n+1
|x|
=
lim
=
=
lim
= lim
n
n
n
n+1
n→∞
n→∞
n→∞
n→∞
nx /2
nx 2
2n
2
n
2
lim
By ratio test we get that when |x|/2 < 1 then the series converges,
that is when |x| < 2 then the series converges. Therefore the radius of
convergence is 2.
(4) (a) Note that
∞
X
1
xn =
.
1−x
n=0
4
Subbing in −x2 for x we get
∞
X
(−1)n x2n =
n=0
1
.
1 + x2
Integrating both sides we get
∞ Z
∞
X
X
(−1)n x2n+1
n 2n
= tan−1 x) + C
(−1) x dx =
(−1)n x2n dx = C 0 +
2n
+
1
n=0
n=0
n=0
Z X
∞
Therefore we get
tan−1 x = K +
∞
X
(−1)n x2n+1
.
2n + 1
n=0
Subbing in x = 0 we get tan−1 x = 0 = K. Therefore
tan−1 x =
∞
X
(−1)n x2n+1
,
2n + 1
n=0
which is the desired result.
(b) Subbing in x = 1 we get
tan−1 1 =
∞
X
π
(−1)n
1 1 1
=
= 1 − + − + ···
4
2n + 1
3 5 7
n=0
(5) (a) From class we know that
sin(x) =
∞
X
(−1)n x2n+1
.
(2n + 1)!
n=0
Subbing in x2 for x we get
sin(x2 ) =
∞
X
(−1)n x4n+2
.
(2n + 1)!
n=0
Integrating both sides we get
Z 1
Z 1X
∞
(−1)n x4n+2
dx
sin(x2 )dx =
0
0 n=0 (2n + 1)!
∞ Z 1
X
(−1)n x4n+2
=
(2n + 1)!
n=0 0
1
∞
X
(−1)n x4n+3 =
(4n + 3)(2n + 1)! n=0
∞
X
x=0
(−1)n
=
.
(4n + 3)(2n + 1)!
n=0
PN
P∞
(−1)n
(b) Notice that |J − n=0 (4n+3)(2n+1)!
| = | n=N +1
this is an alternating series we get
|
∞
X
n=N +1
(−1)n
(4n+3)(2n+1)! |.
(−1)n
1
|<
(4n + 3)(2n + 1)!
(4N + 7)(2N + 3)!
Since
5
If (4N + 7)(2N + 3)! > 10 then the error will be less than 0.1. Note
that when N = 0 we get 7 · 6 = 42 > 10. Therefore J ' 31 with error
less than 0.1.
0
(6) (a) Rf (x) = (cos(x), − sin(x), 1)
x
(b) 0 f (t)dt = (− cos(x), sin(x), x2 /2)
q
√
0
(c) ||f (x)|| = cos2 (x) + sin2 (x) + 1 = 2.