﻿ The Integral Test / Alternating / P-series Worksheet | Problems & Solutions

# The Integral Test / Alternating / P-series Worksheet

The Integral Test / Alternating / P-series Worksheet
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1.
Use the Alternating series test to determine whether the series $\sum _{\mathrm{n=1}}^{\infty }$ $\frac{{\left(-1\right)}^{n}{2}^{-n}}{{n}^{2}+1}$ is convergent or divergent.
 a. cannot be determined b. divergent c. convergent

#### Solution:

Let n=1 an = n=1 (-1)n2-nn2+1
[Original Series.]

Let bn = 2-nn2+1 ≥ 0 for all n.

limn bn = limn 2-nn2+1

= limn 2-n(-ln 2)2n
[Use L'Hospital's rule.]

= limn 2-n(ln 2)22 = 0
[Use L'Hospital's rule.]

Check the sequence {bn} = {2-nn2+1} is decreasing or not.

n2 + 1 < (n + 1)2 + 1 for all n.

1n2+1 > 1(n+1)2+1 for all n.
[If a < b then 1a > 1b.]

2-nn2+1 > 2-n(n+1)2+1 > 2-(n+1)(n+1)2+1 for all n.

So, bn > bn+1 for all n.

That is limn bn = 0 and {bn} is a decreasing sequence, so by Alternating Series Test n=1 (- 1)n bn is convergent.

So, n=1 (-1)n2-nn2+1 is convergent.

2.
Use the Integral Test to determine whether the series $\sum _{\mathrm{n=1}}^{\infty }$ $\frac{3{n}^{2}}{\sqrt{{n}^{3}+8}}$ is convergent or divergent.
 a. oscillating b. divergent c. cannot be determined d. convergent

#### Solution:

f(x) = 3x2x3+8

Since f is positive, continuous and decreasing for x ≥ 1, Integral Test is applicable.

1 3x2x3+8 dx = limR 1R 3x2x3+8 dx

= limR (2x3+8)1R
[Integrate.]

= limR (2R3+8-2(1)3+8)
[Apply the limits of integration.]

= ∞
[Evaluate the limit.]

Therefore, the series diverges.
[n=1 an and 1 f(x) dx, either both converge or both diverge if f is positive, continuous and decreasing for x ≥ 1, where an = f(n).]

So, the series n=1 3n2n3+8 is divergent.

3.
Use the Integral Test to determine whether the series $\sum _{\mathrm{n=2}}^{\infty }$ $\frac{1}{{n}^{2}+4}$ is convergent or divergent.
 a. oscillating b. cannot be determined c. convergent d. divergent

#### Solution:

f(x) = 1x2+4

Since f is positive, continuous and decreasing for x ≥ 2, Integral Test is applicable.

2 1x2+4 dx = limR 2R 1x2+4 dx

= limR 2R 1x2+(2)2 dx

= limR (1 / 2 tan-1(x2))2R
[Integrate.]

= limR (1 / 2 tan-1(R2) - 1 / 2 tan-1(22))
[Apply the limits of integration.]

= π4 - π8

= π8
[Evaluate the limit.]

Therefore, the series converges.
[n=2 an and 2 f(x) dx, either both converge or both diverge if f is positive, continuous and decreasing for x ≥ 2, where an = f(n).]

So, the series n=2 1n2+4 is convergent.

4.
Use the Integral Test to determine whether the series $\sum _{\mathrm{n=1}}^{\infty }$ is convergent or divergent.
 a. cannot be determined b. convergent c. divergent d. oscillating

#### Solution:

f(x) = (ln x)5x

Since f is positive, continuous and decreasing for x ≥ 1, Integral Test is applicable.

1 (ln x)5x dx = limR 1R (ln x)5x dx

= limR 1R 1x(ln x)5 dx

= limR ((ln x)66)1R
[Integrate.]

= limR ((ln R)66 - (ln 1)66)
[Apply the limits of integration.]

= ∞
[Evaluate the limit.]

Therefore, the series diverges.
[n=1 an and 1 f(x) dx, either both converge or both diverge if f is positive, continuous and decreasing for x ≥ 1, where an = f(n).]

So, the series n=1 (ln n)5n is divergent.

5.
Use the Integral Test to determine whether the series $\sum _{\mathrm{n=1}}^{\infty }$ 9$n$3 ${e}^{{-n}^{4}}$ is convergent or divergent.
 a. cannot be determined b. divergent c. convergent

#### Solution:

f(x) = 9x3 e-x4

Since f is positive, continuous and decreasing for x ≥ 1, Integral Test is applicable.

1 9x3 e-x4 dx = limR 1R 9x3 e-x4 dx

= limR (- 9 / 4 e-x4)1R
[Integrate.]

= limR [- 9 / 4 e-R4 - (- 9 / 4e-1)]
[Apply the limits of integration.]

= 94e
[Evaluate the limit.]

Therefore, the series converges.
[n=1 an and 1 f(x) dx, either both converge or both diverge if f is positive, continuous and decreasing for x ≥ 1, where an = f(n).]

So, the series n=1 9n3 e-n4 is convergent.

6.
Use the Integral Test to determine whether the series $\sum _{\mathrm{n=3}}^{\infty }$ $\frac{n}{{\left(n+2\right)}^{\frac{5}{2}}}$ is convergent or divergent.
 a. cannot be determined b. convergent c. divergent

#### Solution:

f(x) = x(x+2)52

Since f is positive, continuous and decreasing for x ≥ 3, Integral Test is applicable.

3 x(x+2)52 dx = limR 3R x(x+2)52 dx

Let x + 2 = u, dx = du

At x = 3, u = 5 and x = R, u = R + 2

limR 3R x(x+2)52 dx = limR 5R+2 (u-32 - 2u-52) du

= limR [(- 2(R+2)-12 + 4 / 3 (R+2)-32) - (- 2(5)-12 + 4 / 3 (5)-32)]
[Integrate and apply the limits of integration.]

= 26575

Therefore, the series converges.
[n=3 an and 3 f(x) dx, either both converge or both diverge if f is positive, continuous and decreasing for x ≥ 3, where an = f(n).]

So, the series n=3 n(n+2)52 is convergent.

7.
Use the $p$-Series Test to determine whether the series $\sum _{\mathrm{n=1}}^{\infty }$ $\frac{1}{{n}^{3}}$ is convergent or divergent.
 a. divergent b. cannot be determined c. convergent

#### Solution:

n=1 1n3
[Write the series.]

Comparing with n=1 1np, p = 3 > 1, the series is convergent.
[n=1 1np is convergent if p > 1 and is divergent if p ≤ 1.]

So, the series n=1 1n3 is convergent.

8.
Use the $p$-Series Test to determine whether the series $\sum _{\mathrm{n=1}}^{\infty }$ $\frac{1}{{n}^{-2}}$ is convergent or divergent.
 a. divergent b. convergent c. cannot be determined

#### Solution:

n=1 1n-2
[Write the series.]

Comparing with n=1 1np, p = - 2 < 1, therefore the series is divergent.
[n=1 1np is convergent if p > 1 and is divergent if p ≤ 1.]

So, the series n=1 1n-2 is divergent.

9.
Use the $p$-Series Test to determine whether the series $\sum _{\mathrm{n=1}}^{\infty }$ $\frac{1}{{n}^{-\left(\frac{1}{4}\right)}}$ is convergent or divergent.
 a. divergent b. cannot be determined c. convergent

#### Solution:

n=1 1n-(14)
[Write the series.]

Comparing with n=1 1np, p = - 1 / 4 < 1, therefore the series is divergent.
[n=1 1np is convergent if p > 1 and is divergent if p ≤ 1.]

So, the series n=1 1n-(14) is divergent.

10.
Use the $p$-Series Test to determine whether the series $\sum _{\mathrm{n=1}}^{\infty }$ $\frac{1}{\sqrt[3]{{n}^{4}}}$ is convergent or divergent.
 a. divergent b. cannot be determined c. convergent

#### Solution:

n=1 1n43
[Write the series.]

Comparing with n=1 1np, p = 4 / 3 > 1, therefore the series is convergent.
[n=1 1np is convergent if p > 1 and is divergent if p ≤ 1.]

So, the series n=1 1n43 is convergent.