Math 2260 Exam #3 Solutions
1. Consider the sequence ( a n
)
∞ n =1 where a n
= 3 n sin(2 /n ) .
Does this sequence converge or diverge? If it converges, what is the limit of the sequence?
Answer: We want to compute the limit lim n →∞
3 n sin(2 /n ) .
Now, this is in the indeterminate form ∞ · 0, so my goal is to re-write and use L’Hˆ notice that the above limit is equal to lim x →∞
3 x sin(2 /x ) = lim x →∞
3 sin(2 /x )
.
1 /x
This is now in the indeterminate form 0 / 0, so we can apply L’Hˆ limit is equal to
− 2 x 2 lim x →∞
3 cos(2
−
/x
1
)
/x 2
·
= lim x →∞
3 · 2 cos(2 /x ) = 3 · 2 · 1 = 6 .
Therefore, the given sequence converges to 6.
2.
(a) Does the series
∞
X n n =2 ln( n ) converge or diverge?
Answer: Notice that the numerator is getting big much faster than the denominator, so we should expect this series to fail the n th term test. Indeed, n lim n →∞ ln( n ) x
= lim x →∞ ln( x )
1
= lim x →∞ 1 /x
= lim x →∞ x = ∞ ,
Therefore, the series fails the n th term test, so it must diverge.
(b) Does your answer to part (a) change if a factor of ( − 1) n
Answer: Inserting a factor of ( − 1) n going to zero. Indeed, is inserted into the series?
won’t change the fact that the terms in the series aren’t lim n →∞
( − 1) n n ln( n )
= lim x →∞
( − 1) n x ln( x )
= lim x →∞
( − 1) n
1
1 /x
= lim x →∞
( − 1) n x, which doesn’t even exist. Therefore, the series again fails the n th term test, so it diverges.
3. Does the series
∞
X
( − 1) n n =2 n [ln( n )] 2 converge absolutely, converge conditionally, or diverge?
Answer: Notice, first of all, that the terms of the series satisfy the hypotheses of the Alternating
Series Test, so the series must converge. It remains only to determine whether it converges absolutely or conditionally.
1

To test absolute convergence, consider the series of absolute values
∞
X
( − 1) n n [ln( n )] 2 n =2
=
∞
X n =2
1 n [ln( n )] 2
.
The easiest way to check whether this series converges is to use the Integral Test:
Z
∞
2
1 x [ln( x )] 2 dx = lim b →∞
Z b
2
1 x [ln( x )] 2 dx.
Now, let u = ln( x ). Then du =
1 x dx and the above integral becomes lim b →∞
Z ln( b ) ln(2)
1 u 2 du = lim b →∞
Z ln( b ) u
− 2 ln(2) du
= lim b →∞
= lim b →∞
1
= 0 + ln(2)
=
1 ln(2)
.
− 1 u ln( b ) ln(2)
− 1 ln( b )
+
1 ln(2)
Therefore, the improper integral converges, so the Integral Test implies that the series also converges.
We conclude that the original series must converge absolutely.
4. Find the interval of convergence of the series
∞
X n =5
(2 x − 3)
√ n − 4 n
.
Answer: First, use the Ratio Test on the absolute values: lim n →∞
(2 x − 3) n +1
√ n − 3
(2 x − 3) n n − 4
= lim n →∞
(2 x − 3) n +1
(2 x − 3) n
·
√ n − 4
√ n − 3
= lim n →∞
| 2 x − 3 | r n − 4 n − 3
= | 2 x − 3 | · 1
= | 2 x − 3 | .
Therefore, the series definitely converges absolutely when | 2 x − 3 | < 1, which is to say when
− 1 < 2 x − 3 < 1 .
Now we check the endpoints. When 2 x − 3 = 1, the series becomes
∞
X n =5
1 n
√ n − 4
=
∞
X n =5
1
√ n − 4
.
2

But now
√ n − 4 <
√ n , so √
1 n − 4
> √ n
. Since we know that the series conclude that the series diverges by the Direct Comparison Test.
P
∞ n =5
√ n diverges, we can
On the other hand, when 2 x − 3 = − 1, the series becomes
∞
X n =5
( − 1) n
√ n − 4
, which satisfies the hypotheses of the Alternating Series Test and hence converges.
Therefore, we see that the given series converges for
− 1 ≤ 2 x − 3 < 1 or
2 ≤ 2 x < 4 or
1 ≤ x < 2 .
Hence, the series converges precisely on the interval [1 , 2).
5. Find the Taylor series centered at x = 1 / 3 for the function f ( x ) = e
− 3 x
.
You will certainly get partial credit for giving the first few terms, but for full credit you should give the general form of the series (i.e., in sigma notation).
Answer: First of all, we’ll need a bunch of derivatives of f : f
0
( x ) = − 3 e
− 3 x f
00
( x ) = 9 e
− 3 x f
000
( x ) = − 27 e
− 3 x
.
..
f
( n )
( x ) = ( − 1) n
3 n e
− 3 x
.
Therefore, f (1 / 3) = e
− 1
=
1 e f
0
(1 / 3) = − 3 e
− 1
=
− 3 e f f
00
000
(1 / 3) = 9 e
− 1
(1 / 3) = − 27 e
=
9 e
− 1
=
− 27 e
.
..
f
( n )
(1 / 3) = ( − 1) n
3 n e
− 1
=
( − 1) n 3 n
.
e
Therefore, the Taylor series is
∞
X n =0 f
( n )
(1 / 3)
( x − 1 / 3) n n !
=
∞
X n =0
( − 1) n
3 n
( x − 1 / 3) n
.
e · n !
3