Using the root test when the limit does not exist
up vote
7
down vote
favorite
I used the root test for the series
$$
sum_{n=1}^{infty} left(frac{cos n}{2}right)^n.
$$
I showed that
$$
0 le left|frac{cos(n)}{2}right| le frac{1}{2} implies lim_{ntoinfty}left|frac{cos(n)}{2}right| le frac{1}{2} < 1.
$$
By the root test, the series converges absolutely. My professor told me that the flaw here is that the limit above does not exist. I agree the limit does not exist because $lvertfrac{cos n}{2}rvert$ oscillates between $0$ and $frac{1}{2}$. However, I fail to see why my argument does not work here. She suggested that I use the comparison test and compare the series with $sum_{n=1}^{infty} left(frac{1}{2}right)^n$. By the comparison test, the original series converges absolutely. Is it a coincidence that the "pseudo" root test I used yielded the same answer as the comparison test? Can we say that if $lvert a_nrvert^{frac{1}{n}}<1$, then $sum_{n=1}^{infty} a_n$ converges absolutely?
I appreciate any help on this.
calculus limits convergence absolute-convergence
New contributor
add a comment |
up vote
7
down vote
favorite
I used the root test for the series
$$
sum_{n=1}^{infty} left(frac{cos n}{2}right)^n.
$$
I showed that
$$
0 le left|frac{cos(n)}{2}right| le frac{1}{2} implies lim_{ntoinfty}left|frac{cos(n)}{2}right| le frac{1}{2} < 1.
$$
By the root test, the series converges absolutely. My professor told me that the flaw here is that the limit above does not exist. I agree the limit does not exist because $lvertfrac{cos n}{2}rvert$ oscillates between $0$ and $frac{1}{2}$. However, I fail to see why my argument does not work here. She suggested that I use the comparison test and compare the series with $sum_{n=1}^{infty} left(frac{1}{2}right)^n$. By the comparison test, the original series converges absolutely. Is it a coincidence that the "pseudo" root test I used yielded the same answer as the comparison test? Can we say that if $lvert a_nrvert^{frac{1}{n}}<1$, then $sum_{n=1}^{infty} a_n$ converges absolutely?
I appreciate any help on this.
calculus limits convergence absolute-convergence
New contributor
3
If you had said that $limsuplimits_{ntoinfty}left|,frac{cos(n)}2,right|lefrac12$, then your statement would be correct, as $limsup$ always exists (though it might be infinite).
– robjohn♦
2 days ago
Thank you. I do not know what lim sup is, and I am reading about it now. I am in my second semester of calculus.
– Htamstudent
2 days ago
$limsup$ and $liminf$ always exist (though each might be infinite). If they are equal, the $lim$ exists; if they are not equal, the $lim$ doesn't exist.
– robjohn♦
2 days ago
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I used the root test for the series
$$
sum_{n=1}^{infty} left(frac{cos n}{2}right)^n.
$$
I showed that
$$
0 le left|frac{cos(n)}{2}right| le frac{1}{2} implies lim_{ntoinfty}left|frac{cos(n)}{2}right| le frac{1}{2} < 1.
$$
By the root test, the series converges absolutely. My professor told me that the flaw here is that the limit above does not exist. I agree the limit does not exist because $lvertfrac{cos n}{2}rvert$ oscillates between $0$ and $frac{1}{2}$. However, I fail to see why my argument does not work here. She suggested that I use the comparison test and compare the series with $sum_{n=1}^{infty} left(frac{1}{2}right)^n$. By the comparison test, the original series converges absolutely. Is it a coincidence that the "pseudo" root test I used yielded the same answer as the comparison test? Can we say that if $lvert a_nrvert^{frac{1}{n}}<1$, then $sum_{n=1}^{infty} a_n$ converges absolutely?
I appreciate any help on this.
calculus limits convergence absolute-convergence
New contributor
I used the root test for the series
$$
sum_{n=1}^{infty} left(frac{cos n}{2}right)^n.
$$
I showed that
$$
0 le left|frac{cos(n)}{2}right| le frac{1}{2} implies lim_{ntoinfty}left|frac{cos(n)}{2}right| le frac{1}{2} < 1.
$$
By the root test, the series converges absolutely. My professor told me that the flaw here is that the limit above does not exist. I agree the limit does not exist because $lvertfrac{cos n}{2}rvert$ oscillates between $0$ and $frac{1}{2}$. However, I fail to see why my argument does not work here. She suggested that I use the comparison test and compare the series with $sum_{n=1}^{infty} left(frac{1}{2}right)^n$. By the comparison test, the original series converges absolutely. Is it a coincidence that the "pseudo" root test I used yielded the same answer as the comparison test? Can we say that if $lvert a_nrvert^{frac{1}{n}}<1$, then $sum_{n=1}^{infty} a_n$ converges absolutely?
I appreciate any help on this.
calculus limits convergence absolute-convergence
calculus limits convergence absolute-convergence
New contributor
New contributor
edited 2 days ago
Viktor Glombik
470220
470220
New contributor
asked 2 days ago
Htamstudent
382
382
New contributor
New contributor
3
If you had said that $limsuplimits_{ntoinfty}left|,frac{cos(n)}2,right|lefrac12$, then your statement would be correct, as $limsup$ always exists (though it might be infinite).
– robjohn♦
2 days ago
Thank you. I do not know what lim sup is, and I am reading about it now. I am in my second semester of calculus.
– Htamstudent
2 days ago
$limsup$ and $liminf$ always exist (though each might be infinite). If they are equal, the $lim$ exists; if they are not equal, the $lim$ doesn't exist.
– robjohn♦
2 days ago
add a comment |
3
If you had said that $limsuplimits_{ntoinfty}left|,frac{cos(n)}2,right|lefrac12$, then your statement would be correct, as $limsup$ always exists (though it might be infinite).
– robjohn♦
2 days ago
Thank you. I do not know what lim sup is, and I am reading about it now. I am in my second semester of calculus.
– Htamstudent
2 days ago
$limsup$ and $liminf$ always exist (though each might be infinite). If they are equal, the $lim$ exists; if they are not equal, the $lim$ doesn't exist.
– robjohn♦
2 days ago
3
3
If you had said that $limsuplimits_{ntoinfty}left|,frac{cos(n)}2,right|lefrac12$, then your statement would be correct, as $limsup$ always exists (though it might be infinite).
– robjohn♦
2 days ago
If you had said that $limsuplimits_{ntoinfty}left|,frac{cos(n)}2,right|lefrac12$, then your statement would be correct, as $limsup$ always exists (though it might be infinite).
– robjohn♦
2 days ago
Thank you. I do not know what lim sup is, and I am reading about it now. I am in my second semester of calculus.
– Htamstudent
2 days ago
Thank you. I do not know what lim sup is, and I am reading about it now. I am in my second semester of calculus.
– Htamstudent
2 days ago
$limsup$ and $liminf$ always exist (though each might be infinite). If they are equal, the $lim$ exists; if they are not equal, the $lim$ doesn't exist.
– robjohn♦
2 days ago
$limsup$ and $liminf$ always exist (though each might be infinite). If they are equal, the $lim$ exists; if they are not equal, the $lim$ doesn't exist.
– robjohn♦
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
We have that
$$ left|left(frac{cos n}{2}right)^nright|le frac1{2^n}$$
and $sum frac1{2^n}$ is a convergent geometric series, we don't need root test here.
Anyway we can also apply root test to the original series in the general form by limsup definition
$$limsup_{nrightarrowinfty}sqrt[n]{left|left(frac{cos n}{2}right)^nright|}=Lle frac12$$
and conclude that the series converges.
add a comment |
up vote
8
down vote
The root test can be used without the sequence having a limit. Precisely,
if there exist $N$ and $c<1$ with $sqrt[n]{|a_n|}le c$ for all $n>N$, then the series $sum_{n=0}^infty a_n$ is absolutely convergent.
Indeed, in this case one can directly compare the series with a convergent geometric series. When $lim_{ntoinfty}sqrt[n]{|a_n|}=l$ exists and is $<1$, then the above criterion applies, because we can take $c=(l+1)/2$.
If you had used the “extended criterion” rather than stating that $lim_{ntoinfty}lvertfrac{cos n}{2}rvertle frac{1}{2}$, you would be right.
Thank you. I just learned the extended criterion from you and robjohn today.
– Htamstudent
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
We have that
$$ left|left(frac{cos n}{2}right)^nright|le frac1{2^n}$$
and $sum frac1{2^n}$ is a convergent geometric series, we don't need root test here.
Anyway we can also apply root test to the original series in the general form by limsup definition
$$limsup_{nrightarrowinfty}sqrt[n]{left|left(frac{cos n}{2}right)^nright|}=Lle frac12$$
and conclude that the series converges.
add a comment |
up vote
6
down vote
accepted
We have that
$$ left|left(frac{cos n}{2}right)^nright|le frac1{2^n}$$
and $sum frac1{2^n}$ is a convergent geometric series, we don't need root test here.
Anyway we can also apply root test to the original series in the general form by limsup definition
$$limsup_{nrightarrowinfty}sqrt[n]{left|left(frac{cos n}{2}right)^nright|}=Lle frac12$$
and conclude that the series converges.
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
We have that
$$ left|left(frac{cos n}{2}right)^nright|le frac1{2^n}$$
and $sum frac1{2^n}$ is a convergent geometric series, we don't need root test here.
Anyway we can also apply root test to the original series in the general form by limsup definition
$$limsup_{nrightarrowinfty}sqrt[n]{left|left(frac{cos n}{2}right)^nright|}=Lle frac12$$
and conclude that the series converges.
We have that
$$ left|left(frac{cos n}{2}right)^nright|le frac1{2^n}$$
and $sum frac1{2^n}$ is a convergent geometric series, we don't need root test here.
Anyway we can also apply root test to the original series in the general form by limsup definition
$$limsup_{nrightarrowinfty}sqrt[n]{left|left(frac{cos n}{2}right)^nright|}=Lle frac12$$
and conclude that the series converges.
edited 2 days ago
answered 2 days ago
gimusi
85.5k74294
85.5k74294
add a comment |
add a comment |
up vote
8
down vote
The root test can be used without the sequence having a limit. Precisely,
if there exist $N$ and $c<1$ with $sqrt[n]{|a_n|}le c$ for all $n>N$, then the series $sum_{n=0}^infty a_n$ is absolutely convergent.
Indeed, in this case one can directly compare the series with a convergent geometric series. When $lim_{ntoinfty}sqrt[n]{|a_n|}=l$ exists and is $<1$, then the above criterion applies, because we can take $c=(l+1)/2$.
If you had used the “extended criterion” rather than stating that $lim_{ntoinfty}lvertfrac{cos n}{2}rvertle frac{1}{2}$, you would be right.
Thank you. I just learned the extended criterion from you and robjohn today.
– Htamstudent
2 days ago
add a comment |
up vote
8
down vote
The root test can be used without the sequence having a limit. Precisely,
if there exist $N$ and $c<1$ with $sqrt[n]{|a_n|}le c$ for all $n>N$, then the series $sum_{n=0}^infty a_n$ is absolutely convergent.
Indeed, in this case one can directly compare the series with a convergent geometric series. When $lim_{ntoinfty}sqrt[n]{|a_n|}=l$ exists and is $<1$, then the above criterion applies, because we can take $c=(l+1)/2$.
If you had used the “extended criterion” rather than stating that $lim_{ntoinfty}lvertfrac{cos n}{2}rvertle frac{1}{2}$, you would be right.
Thank you. I just learned the extended criterion from you and robjohn today.
– Htamstudent
2 days ago
add a comment |
up vote
8
down vote
up vote
8
down vote
The root test can be used without the sequence having a limit. Precisely,
if there exist $N$ and $c<1$ with $sqrt[n]{|a_n|}le c$ for all $n>N$, then the series $sum_{n=0}^infty a_n$ is absolutely convergent.
Indeed, in this case one can directly compare the series with a convergent geometric series. When $lim_{ntoinfty}sqrt[n]{|a_n|}=l$ exists and is $<1$, then the above criterion applies, because we can take $c=(l+1)/2$.
If you had used the “extended criterion” rather than stating that $lim_{ntoinfty}lvertfrac{cos n}{2}rvertle frac{1}{2}$, you would be right.
The root test can be used without the sequence having a limit. Precisely,
if there exist $N$ and $c<1$ with $sqrt[n]{|a_n|}le c$ for all $n>N$, then the series $sum_{n=0}^infty a_n$ is absolutely convergent.
Indeed, in this case one can directly compare the series with a convergent geometric series. When $lim_{ntoinfty}sqrt[n]{|a_n|}=l$ exists and is $<1$, then the above criterion applies, because we can take $c=(l+1)/2$.
If you had used the “extended criterion” rather than stating that $lim_{ntoinfty}lvertfrac{cos n}{2}rvertle frac{1}{2}$, you would be right.
answered 2 days ago
egreg
173k1383197
173k1383197
Thank you. I just learned the extended criterion from you and robjohn today.
– Htamstudent
2 days ago
add a comment |
Thank you. I just learned the extended criterion from you and robjohn today.
– Htamstudent
2 days ago
Thank you. I just learned the extended criterion from you and robjohn today.
– Htamstudent
2 days ago
Thank you. I just learned the extended criterion from you and robjohn today.
– Htamstudent
2 days ago
add a comment |
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3
If you had said that $limsuplimits_{ntoinfty}left|,frac{cos(n)}2,right|lefrac12$, then your statement would be correct, as $limsup$ always exists (though it might be infinite).
– robjohn♦
2 days ago
Thank you. I do not know what lim sup is, and I am reading about it now. I am in my second semester of calculus.
– Htamstudent
2 days ago
$limsup$ and $liminf$ always exist (though each might be infinite). If they are equal, the $lim$ exists; if they are not equal, the $lim$ doesn't exist.
– robjohn♦
2 days ago