Answers - Bruce E. Shapiro

Math 150B - Test 4 - Solutions
1. Determine if the following series converges or if it diverges, clearly explaining the
∞∑ 1
reasoning behind your conclusion:
n √ ln n
Using the integral test
∫ ∞
2
n=2
∫
dx
x=∞
x √ ln x = dx
√ where u = ln x, and du = dx x x
x=2
= 2 √ u ∣ ∣ x=∞
x=2 = 2 (√
ln ∞ −
√
ln 2
)
→ ∞ =⇒ DIVERGES
2. Determine if the following series converges or if it diverges, clearly explaining the
∞∑ n 2n
reasoning behind your conclusion:
(1 + 2n 2 ) n
Using the root test
n=1
( )
(|a n |) 1/n n 2n 1/n
=
= n2
(1 + 2n 2 ) n 1 + 2n → 1 2 2 as n → ∞
Since 1/2 < 1 the series CONVERGES absolutely.
3. Determine if the following series converges or if it diverges, clearly explaining the
∞∑ n 3
reasoning behind your conclusion:
5 n
n=1
Using the ratio test,
∣ ∣ a n+1 ∣∣∣ ∣ =
(n + 1) 3 ∣∣∣
a n
∣ × 5n
=
5 n+1 n 3
∣
(n + 1) 3
n 3 ∣ ∣∣∣
∣ ∣∣∣ 5 n
5 × 5 n ∣ ∣∣∣
→ 1 5 < 1 as n → ∞
Since the ratio converges to a limit that is less the 1 the series CONVERGES by the
ratio test.
4. Determine if the following series converges or if it diverges, clearly explaining the
∞∑ (−1) n
reasoning behind your conclusion: √ n + 1
This is an alternating series so apply the alternating series test.
n=1
(a) Since √ n < √ n + 1 =⇒ 1 √ n + 1
< 1 √ n
=⇒ |a n+1 | < |a n | the series is monotonically
decreasing.
(b) Since 1/ √ n + 1 → 0 as n → ∞ then a n → 0 as n → ∞.
Hence the series CONVERGES by the alternating series test.
1

Solutions
5. Find a power series for f(x) = x2
. What are its radius and interval of convergence?
1 + x
Using the formula for the sum of a geometric series, the power series is
x 2
1 + x = x2 ( 1 − x + x 2 − x 3 + · · · ) = x 2 − x 3 + x 4 − x 5 + · · ·
The formula for the geometric series holds for |x| < 1. Hence the radius of convergence
is 1 and the interval of convergence is −1 < x < 1.
6. How many terms of the series
∞∑ (−1) n+1
n=1
accuracy of of better than 0.00001?
n 5
This is an alternating series so we require
|a n+1 | =
are needed to compute the sum to an
1
(n + 1) 5 < 0.00001 = 10−5 = 1
10 5 =⇒ (n + 1)5 > 10 5 =⇒ n + 1 > 10 =⇒ n > 9
7. Find the radius and interval of convergence of the power series
∞∑
(−1) n xn
n 2 5 . n
Use the ratio test:
∣ ∣ ∣ a n+1 ∣∣∣ ∣ =
x n+1
a n
∣(n + 1) 2 5 × n2 5 n ∣∣∣ =
n 2 x ∣∣∣ n+1 x n ∣ → |x|
5(n + 1) 2 5 as n → ∞
We need this limit to be < 1 or |x| < 5 =⇒ −5 < x < 5. The RADIUS OF
CONVERGENCE is 5.
∞∑
Checking the endpoints: at x = −5, the sum is (−1) n (−5)n
n 2 5 = ∑ ∞
1
which is a
n n 2
n=1
n=1
p-series with p=2, so it CONVERGES.
∞∑
At x = 5, the series is (−1) n (5)n
n 2 5 = ∑ ∞
(−1) n
which CONVERGES ABSOLUTELY
n n 2
n=1
n=1
(because the series in the previous sentence converges) hence it CONVERGES.
Thus the INTERVAL of CONVERGENCE is −5 ≤ x ≤ 5.
8. Write the repeating decimal 4.17326326326 as a fraction of integers.
x = 4.17326326
1000x = 4173.26326326
999x = 4169.09
x = 4169.09
999
= 416909
99900
n=1
2

∞∑ (−3) n−1
9. Find the sum of the series
n=1
2 3n
Writing out the first few terms, the series is
Solutions
S = 1 8 + −3
8 2 + (−3)2
8 3 + (−3)3
8 4 + · · ·
this is a geometric series with a = 1/8 and r = −3/8 so the sum is
S =
10. Find the sum of the series
a
1 − r = 1/8
1 − (−3/8) = 1/8
11/8 = 1 11
∞∑ (
tan −1 (n + 1) − tan −1 n )
n=1
Writing out the first we terms we see that the series telescopes:
S = (tan −1 2 − tan −1 1) + (tan −1 3 − tan −1 2) + (tan −1 4 − tan −1 3)+
(tan −1 5 − tan −1 4) + (tan −1 6 − tan −1 5) + · · ·
= − tan −1 1
= − π 4
3