For all a > 0 and b> 1, the inequalities In (n) ≤ nº, nº N. (Express numbers in exact form. Use symbolic notation and fractions where needed.) inequality: In (N)¹ ≤N Draw a conclusion about the convergence of O It is not possible to use the Direct Comparison Test to determine the convergence or divergence of 00 Because n= n=1 Because is a harmonic series, it diverges. Therefore, P ∞ 1 Σ – converges, Σ n N=1 (In (n))" 1 (In (n))7 ∞ 1 (In (n))" also converges by the Direct Comparison Test. n=1 00 (In (n))7 diverges by the Direct Comparison Test.
For all a > 0 and b> 1, the inequalities In (n) ≤ nº, nº N. (Express numbers in exact form. Use symbolic notation and fractions where needed.) inequality: In (N)¹ ≤N Draw a conclusion about the convergence of O It is not possible to use the Direct Comparison Test to determine the convergence or divergence of 00 Because n= n=1 Because is a harmonic series, it diverges. Therefore, P ∞ 1 Σ – converges, Σ n N=1 (In (n))" 1 (In (n))7 ∞ 1 (In (n))" also converges by the Direct Comparison Test. n=1 00 (In (n))7 diverges by the Direct Comparison Test.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![80
F3
For all a > 0 and b > 1, the inequalities In (n) ≤ nº, nº <b" are true for n sufficiently large (this can be proved using
L'Hôpital's Rule).
Let N be this sufficiently large n. Write an inequality comparing (In (n))7 and n for n > N.
(Express numbers in exact form. Use symbolic notation and fractions where needed.)
$
R
F
inequality: In (N) ≤N
Draw a conclusion about the convergence of
Because
O
It is not possible to use the Direct Comparison Test to determine the convergence or divergence of
1
n=1
(In (n))
diverges by the Direct Comparison Test.
a
1
Because is a harmonic series, it diverges. Therefore, Σ
n
n=1
F4
%
5
n=1
n=1
T
G
00
· converges, Σ
n
n=1
O
F5
6
Y
F6
H
1
(In (n))7
C
00
&
n=1
7
(In (n))7
also converges by the Direct Comparison Test.
F7
J
*
8
DII
F8
1
(In (n))
I
K
(
9
DD
F9
O
O
L
Question Source: Rogawski 4e Calculus Early Transcendentals P
F
F10
P
F11
{
[
F12
11
1
]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F497e105d-b0e8-47a3-88d6-3cb52a491c4d%2F694f4bfe-06e1-426a-b271-06347e5b3de6%2Ff807g8d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:80
F3
For all a > 0 and b > 1, the inequalities In (n) ≤ nº, nº <b" are true for n sufficiently large (this can be proved using
L'Hôpital's Rule).
Let N be this sufficiently large n. Write an inequality comparing (In (n))7 and n for n > N.
(Express numbers in exact form. Use symbolic notation and fractions where needed.)
$
R
F
inequality: In (N) ≤N
Draw a conclusion about the convergence of
Because
O
It is not possible to use the Direct Comparison Test to determine the convergence or divergence of
1
n=1
(In (n))
diverges by the Direct Comparison Test.
a
1
Because is a harmonic series, it diverges. Therefore, Σ
n
n=1
F4
%
5
n=1
n=1
T
G
00
· converges, Σ
n
n=1
O
F5
6
Y
F6
H
1
(In (n))7
C
00
&
n=1
7
(In (n))7
also converges by the Direct Comparison Test.
F7
J
*
8
DII
F8
1
(In (n))
I
K
(
9
DD
F9
O
O
L
Question Source: Rogawski 4e Calculus Early Transcendentals P
F
F10
P
F11
{
[
F12
11
1
]
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