Mixing
Determine whether the given series is absolutely convergent or conditionally convergent
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Consider the series
$$sum_n=1^infty logleft(1+frac1right).$$
Determine whether it converges absolutely or conditionally.
I am trying to apply Cauchy condensation test, but I am not sure whether the given series is non-increasing or not.
trigonometry summation logarithms absolute-convergence conditional-convergence
edited Jul 31 at 4:35
Math Lover
12.2k21132
asked Jul 31 at 3:55
blue boy
528211
add a comment |Â
up vote
0
down vote
favorite
Consider the series
$$sum_n=1^infty logleft(1+frac1right).$$
Determine whether it converges absolutely or conditionally.
I am trying to apply Cauchy condensation test, but I am not sure whether the given series is non-increasing or not.
trigonometry summation logarithms absolute-convergence conditional-convergence
edited Jul 31 at 4:35
Math Lover
12.2k21132
asked Jul 31 at 3:55
blue boy
528211
Can you prove it converges at all?
– Steve D
Jul 31 at 3:59
It can't converge. The limit as $n to infty$ doesn't equal zero. In fact, the minimum of the equation being summed up is $log(2) approx .30103$. Therefore, the series diverges.
– Christopher Marley
Jul 31 at 4:11
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the series
$$sum_n=1^infty logleft(1+frac1right).$$
Determine whether it converges absolutely or conditionally.
I am trying to apply Cauchy condensation test, but I am not sure whether the given series is non-increasing or not.
trigonometry summation logarithms absolute-convergence conditional-convergence
edited Jul 31 at 4:35
Math Lover
12.2k21132
asked Jul 31 at 3:55
blue boy
528211
Consider the series
$$sum_n=1^infty logleft(1+frac1right).$$
Determine whether it converges absolutely or conditionally.
I am trying to apply Cauchy condensation test, but I am not sure whether the given series is non-increasing or not.
trigonometry summation logarithms absolute-convergence conditional-convergence
edited Jul 31 at 4:35
Math Lover
12.2k21132
asked Jul 31 at 3:55
blue boy
528211
edited Jul 31 at 4:35
Math Lover
12.2k21132
edited Jul 31 at 4:35
Math Lover
12.2k21132
edited Jul 31 at 4:35
Math Lover
12.2k21132
12.2k21132
asked Jul 31 at 3:55
blue boy
528211
asked Jul 31 at 3:55
blue boy
528211
asked Jul 31 at 3:55
blue boy
528211
528211
Can you prove it converges at all?
– Steve D
Jul 31 at 3:59
It can't converge. The limit as $n to infty$ doesn't equal zero. In fact, the minimum of the equation being summed up is $log(2) approx .30103$. Therefore, the series diverges.
– Christopher Marley
Jul 31 at 4:11
add a comment |Â
Can you prove it converges at all?
– Steve D
Jul 31 at 3:59
It can't converge. The limit as $n to infty$ doesn't equal zero. In fact, the minimum of the equation being summed up is $log(2) approx .30103$. Therefore, the series diverges.
– Christopher Marley
Jul 31 at 4:11
Can you prove it converges at all?
– Steve D
Jul 31 at 3:59
Can you prove it converges at all?
– Steve D
Jul 31 at 3:59
It can't converge. The limit as $n to infty$ doesn't equal zero. In fact, the minimum of the equation being summed up is $log(2) approx .30103$. Therefore, the series diverges.
– Christopher Marley
Jul 31 at 4:11
It can't converge. The limit as $n to infty$ doesn't equal zero. In fact, the minimum of the equation being summed up is $log(2) approx .30103$. Therefore, the series diverges.
– Christopher Marley
Jul 31 at 4:11
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
No, the term $logleft(1+frac1right)$ is not decreasing, but since $|sin(x)|leq 1$, it follows that
$$ logleft(1+frac1right)geq logleft(1+frac11right).$$
Can you take it from here?
answered Jul 31 at 4:20
Robert Z
83.6k954122
And log(1+1/n ) diverges . It can be checked by comparing it with 1/n.
– blue boy
Jul 31 at 4:27
@blueboy Actually a stronger estimate holds.
– Robert Z
Jul 31 at 5:11
Yes. Thanks ...
– blue boy
Jul 31 at 5:17
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
No, the term $logleft(1+frac1right)$ is not decreasing, but since $|sin(x)|leq 1$, it follows that
$$ logleft(1+frac1right)geq logleft(1+frac11right).$$
Can you take it from here?
answered Jul 31 at 4:20
Robert Z
83.6k954122
And log(1+1/n ) diverges . It can be checked by comparing it with 1/n.
– blue boy
Jul 31 at 4:27
@blueboy Actually a stronger estimate holds.
– Robert Z
Jul 31 at 5:11
Yes. Thanks ...
– blue boy
Jul 31 at 5:17
add a comment |Â
up vote
2
down vote
accepted
No, the term $logleft(1+frac1right)$ is not decreasing, but since $|sin(x)|leq 1$, it follows that
$$ logleft(1+frac1right)geq logleft(1+frac11right).$$
Can you take it from here?
answered Jul 31 at 4:20
Robert Z
83.6k954122
And log(1+1/n ) diverges . It can be checked by comparing it with 1/n.
– blue boy
Jul 31 at 4:27
@blueboy Actually a stronger estimate holds.
– Robert Z
Jul 31 at 5:11
Yes. Thanks ...
– blue boy
Jul 31 at 5:17
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
No, the term $logleft(1+frac1right)$ is not decreasing, but since $|sin(x)|leq 1$, it follows that
$$ logleft(1+frac1right)geq logleft(1+frac11right).$$
Can you take it from here?
answered Jul 31 at 4:20
Robert Z
83.6k954122
No, the term $logleft(1+frac1right)$ is not decreasing, but since $|sin(x)|leq 1$, it follows that
$$ logleft(1+frac1right)geq logleft(1+frac11right).$$
Can you take it from here?
answered Jul 31 at 4:20
Robert Z
83.6k954122
edited Jul 31 at 5:10
answered Jul 31 at 4:20
Robert Z
83.6k954122
answered Jul 31 at 4:20
Robert Z
83.6k954122
answered Jul 31 at 4:20
Robert Z
83.6k954122
83.6k954122
And log(1+1/n ) diverges . It can be checked by comparing it with 1/n.
– blue boy
Jul 31 at 4:27
@blueboy Actually a stronger estimate holds.
– Robert Z
Jul 31 at 5:11
Yes. Thanks ...
– blue boy
Jul 31 at 5:17
add a comment |Â
And log(1+1/n ) diverges . It can be checked by comparing it with 1/n.
– blue boy
Jul 31 at 4:27
@blueboy Actually a stronger estimate holds.
– Robert Z
Jul 31 at 5:11
Yes. Thanks ...
– blue boy
Jul 31 at 5:17
And log(1+1/n ) diverges . It can be checked by comparing it with 1/n.
– blue boy
Jul 31 at 4:27
And log(1+1/n ) diverges . It can be checked by comparing it with 1/n.
– blue boy
Jul 31 at 4:27
@blueboy Actually a stronger estimate holds.
– Robert Z
Jul 31 at 5:11
@blueboy Actually a stronger estimate holds.
– Robert Z
Jul 31 at 5:11
Yes. Thanks ...
– blue boy
Jul 31 at 5:17
Yes. Thanks ...
– blue boy
Jul 31 at 5:17
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2867641%2fdetermine-whether-the-given-series-is-absolutely-convergent-or-conditionally-con%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Can you prove it converges at all?
– Steve D
Jul 31 at 3:59
It can't converge. The limit as $n to infty$ doesn't equal zero. In fact, the minimum of the equation being summed up is $log(2) approx .30103$. Therefore, the series diverges.
– Christopher Marley
Jul 31 at 4:11