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Convergence of $sum_n=1^inftybigg((2lambda)^3n lnbig(cos(n^lambda)big)bigg)$, for $lambdaleqslant0, lambdainmathbbR$
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Study the convergence of the following series for $lambdaleqslant0, lambdainmathbbR$
$$sum_n=1^inftybigg((2lambda)^3n lnbig(!cos(n^lambda)big)bigg)$$
I observed that if $-frac12ltlambdalt0$ then $(2lambda)^3n$ is the geometric series with ratio $-1lt r lt0$ that converges, whereas for $lambdalt -frac12$ we have $rlt-1$ that diverges. In particular when $lambdalt -frac12$, the $(2lambda)^3n$ is an oscillating series.
I am not sure how to handle $b_n=lnbig(!cos(n^lambda)big)$ which tend to $0$ for $nmapstoinfty$. Giving $k=-lambdagt0$, can I say that $b_n= Obigg(frac1n^kbigg)$ for $n mapsto infty$? Or am I wrong, having lost too much informations?
real-analysis sequences-and-series convergence
edited 14 hours ago
Steven Stadnicki
40.1k765119
asked 14 hours ago
F.inc
1488
add a comment |Â
up vote
2
down vote
favorite
Study the convergence of the following series for $lambdaleqslant0, lambdainmathbbR$
$$sum_n=1^inftybigg((2lambda)^3n lnbig(!cos(n^lambda)big)bigg)$$
I observed that if $-frac12ltlambdalt0$ then $(2lambda)^3n$ is the geometric series with ratio $-1lt r lt0$ that converges, whereas for $lambdalt -frac12$ we have $rlt-1$ that diverges. In particular when $lambdalt -frac12$, the $(2lambda)^3n$ is an oscillating series.
I am not sure how to handle $b_n=lnbig(!cos(n^lambda)big)$ which tend to $0$ for $nmapstoinfty$. Giving $k=-lambdagt0$, can I say that $b_n= Obigg(frac1n^kbigg)$ for $n mapsto infty$? Or am I wrong, having lost too much informations?
real-analysis sequences-and-series convergence
edited 14 hours ago
Steven Stadnicki
40.1k765119
asked 14 hours ago
F.inc
1488
2
Saying that $b_n=O(frac1n^lambda)$ - or even $b_n=O(frac1n^2lambda)$, as in the answer below - isn't actually enough in and of itself, because $O()$ doesn't say anything about how small a term can get. Instead, for $lambdagt-1/2$, you need to bound $ln(cos(n^lambda))$ away from zero, to show that it's not so small that it can 'counteract' the growth of $(2lambda)^3n$.
– Steven Stadnicki
14 hours ago
1
@StevenStadnicki My aim was that of using the convergence/divergence of the harmonic series (maybe Leibnitz and Dirichlet when combined with an oscillating one). Do you find that this idea fails? What would you do? Thank you!
– F.inc
13 hours ago
1
The problem with using the divergence of the geometric series for $lambdalt-1/2$ is that you have to show, not that $ln(cos(n^lambda))$ is small enough, but that it's large enough; in principle it's possible that, for instance, $ln(cos(n^lambda))lt n^-n$, in which case the presence of the geometrically-growing factor of $(2lambda)^3n$ wouldn't be enough to guarantee divergence. This doesn't actually happen, but you have to prove that it doesn't happen.
– Steven Stadnicki
11 hours ago
1
The idea of using the geometric series for comparative purposes is a good one, but you have to make sure that your bounds are correct. (Also, you should make sure to handle $lambda=-1/2$ explicitly as well.)
– Steven Stadnicki
11 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Study the convergence of the following series for $lambdaleqslant0, lambdainmathbbR$
$$sum_n=1^inftybigg((2lambda)^3n lnbig(!cos(n^lambda)big)bigg)$$
I observed that if $-frac12ltlambdalt0$ then $(2lambda)^3n$ is the geometric series with ratio $-1lt r lt0$ that converges, whereas for $lambdalt -frac12$ we have $rlt-1$ that diverges. In particular when $lambdalt -frac12$, the $(2lambda)^3n$ is an oscillating series.
I am not sure how to handle $b_n=lnbig(!cos(n^lambda)big)$ which tend to $0$ for $nmapstoinfty$. Giving $k=-lambdagt0$, can I say that $b_n= Obigg(frac1n^kbigg)$ for $n mapsto infty$? Or am I wrong, having lost too much informations?
real-analysis sequences-and-series convergence
edited 14 hours ago
Steven Stadnicki
40.1k765119
asked 14 hours ago
F.inc
1488
Study the convergence of the following series for $lambdaleqslant0, lambdainmathbbR$
$$sum_n=1^inftybigg((2lambda)^3n lnbig(!cos(n^lambda)big)bigg)$$
I observed that if $-frac12ltlambdalt0$ then $(2lambda)^3n$ is the geometric series with ratio $-1lt r lt0$ that converges, whereas for $lambdalt -frac12$ we have $rlt-1$ that diverges. In particular when $lambdalt -frac12$, the $(2lambda)^3n$ is an oscillating series.
I am not sure how to handle $b_n=lnbig(!cos(n^lambda)big)$ which tend to $0$ for $nmapstoinfty$. Giving $k=-lambdagt0$, can I say that $b_n= Obigg(frac1n^kbigg)$ for $n mapsto infty$? Or am I wrong, having lost too much informations?
real-analysis sequences-and-series convergence
edited 14 hours ago
Steven Stadnicki
40.1k765119
asked 14 hours ago
F.inc
1488
edited 14 hours ago
Steven Stadnicki
40.1k765119
edited 14 hours ago
Steven Stadnicki
40.1k765119
edited 14 hours ago
Steven Stadnicki
40.1k765119
40.1k765119
asked 14 hours ago
F.inc
1488
asked 14 hours ago
F.inc
1488
asked 14 hours ago
F.inc
1488
1488
2
Saying that $b_n=O(frac1n^lambda)$ - or even $b_n=O(frac1n^2lambda)$, as in the answer below - isn't actually enough in and of itself, because $O()$ doesn't say anything about how small a term can get. Instead, for $lambdagt-1/2$, you need to bound $ln(cos(n^lambda))$ away from zero, to show that it's not so small that it can 'counteract' the growth of $(2lambda)^3n$.
– Steven Stadnicki
14 hours ago
1
@StevenStadnicki My aim was that of using the convergence/divergence of the harmonic series (maybe Leibnitz and Dirichlet when combined with an oscillating one). Do you find that this idea fails? What would you do? Thank you!
– F.inc
13 hours ago
1
The problem with using the divergence of the geometric series for $lambdalt-1/2$ is that you have to show, not that $ln(cos(n^lambda))$ is small enough, but that it's large enough; in principle it's possible that, for instance, $ln(cos(n^lambda))lt n^-n$, in which case the presence of the geometrically-growing factor of $(2lambda)^3n$ wouldn't be enough to guarantee divergence. This doesn't actually happen, but you have to prove that it doesn't happen.
– Steven Stadnicki
11 hours ago
1
The idea of using the geometric series for comparative purposes is a good one, but you have to make sure that your bounds are correct. (Also, you should make sure to handle $lambda=-1/2$ explicitly as well.)
– Steven Stadnicki
11 hours ago
add a comment |Â
2
Saying that $b_n=O(frac1n^lambda)$ - or even $b_n=O(frac1n^2lambda)$, as in the answer below - isn't actually enough in and of itself, because $O()$ doesn't say anything about how small a term can get. Instead, for $lambdagt-1/2$, you need to bound $ln(cos(n^lambda))$ away from zero, to show that it's not so small that it can 'counteract' the growth of $(2lambda)^3n$.
– Steven Stadnicki
14 hours ago
1
@StevenStadnicki My aim was that of using the convergence/divergence of the harmonic series (maybe Leibnitz and Dirichlet when combined with an oscillating one). Do you find that this idea fails? What would you do? Thank you!
– F.inc
13 hours ago
1
The problem with using the divergence of the geometric series for $lambdalt-1/2$ is that you have to show, not that $ln(cos(n^lambda))$ is small enough, but that it's large enough; in principle it's possible that, for instance, $ln(cos(n^lambda))lt n^-n$, in which case the presence of the geometrically-growing factor of $(2lambda)^3n$ wouldn't be enough to guarantee divergence. This doesn't actually happen, but you have to prove that it doesn't happen.
– Steven Stadnicki
11 hours ago
1
The idea of using the geometric series for comparative purposes is a good one, but you have to make sure that your bounds are correct. (Also, you should make sure to handle $lambda=-1/2$ explicitly as well.)
– Steven Stadnicki
11 hours ago
2
2
Saying that $b_n=O(frac1n^lambda)$ - or even $b_n=O(frac1n^2lambda)$, as in the answer below - isn't actually enough in and of itself, because $O()$ doesn't say anything about how small a term can get. Instead, for $lambdagt-1/2$, you need to bound $ln(cos(n^lambda))$ away from zero, to show that it's not so small that it can 'counteract' the growth of $(2lambda)^3n$.
– Steven Stadnicki
14 hours ago
Saying that $b_n=O(frac1n^lambda)$ - or even $b_n=O(frac1n^2lambda)$, as in the answer below - isn't actually enough in and of itself, because $O()$ doesn't say anything about how small a term can get. Instead, for $lambdagt-1/2$, you need to bound $ln(cos(n^lambda))$ away from zero, to show that it's not so small that it can 'counteract' the growth of $(2lambda)^3n$.
– Steven Stadnicki
14 hours ago
1
1
@StevenStadnicki My aim was that of using the convergence/divergence of the harmonic series (maybe Leibnitz and Dirichlet when combined with an oscillating one). Do you find that this idea fails? What would you do? Thank you!
– F.inc
13 hours ago
@StevenStadnicki My aim was that of using the convergence/divergence of the harmonic series (maybe Leibnitz and Dirichlet when combined with an oscillating one). Do you find that this idea fails? What would you do? Thank you!
– F.inc
13 hours ago
1
1
The problem with using the divergence of the geometric series for $lambdalt-1/2$ is that you have to show, not that $ln(cos(n^lambda))$ is small enough, but that it's large enough; in principle it's possible that, for instance, $ln(cos(n^lambda))lt n^-n$, in which case the presence of the geometrically-growing factor of $(2lambda)^3n$ wouldn't be enough to guarantee divergence. This doesn't actually happen, but you have to prove that it doesn't happen.
– Steven Stadnicki
11 hours ago
The problem with using the divergence of the geometric series for $lambdalt-1/2$ is that you have to show, not that $ln(cos(n^lambda))$ is small enough, but that it's large enough; in principle it's possible that, for instance, $ln(cos(n^lambda))lt n^-n$, in which case the presence of the geometrically-growing factor of $(2lambda)^3n$ wouldn't be enough to guarantee divergence. This doesn't actually happen, but you have to prove that it doesn't happen.
– Steven Stadnicki
11 hours ago
1
1
The idea of using the geometric series for comparative purposes is a good one, but you have to make sure that your bounds are correct. (Also, you should make sure to handle $lambda=-1/2$ explicitly as well.)
– Steven Stadnicki
11 hours ago
The idea of using the geometric series for comparative purposes is a good one, but you have to make sure that your bounds are correct. (Also, you should make sure to handle $lambda=-1/2$ explicitly as well.)
– Steven Stadnicki
11 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Solution. $blacktriangleleft$ First determine the radius of convergence for
$$
sum y^n log(cos (n^lambda)).
$$
By Cauchy-Hadamard formula, we need to calculate the following:
beginalign*
left| log (cos (n^lambda)) right|^1/n &= exp left( frac 1n log (- log (cos (n^lambda)))right) \
&= exp left( frac 1n log (-log (1 - n^2lambda/2 + o(n^2lambda)))right)\
&= exp left( frac 1n log (n^2lambda + o(n^2lambda))right).
endalign*
For sufficiently large $n$ we can assume that
$$
frac 12 n ^2lambda leqslant n^2lambda + o(n^2lambda) leqslant frac 32 n^2lambda.
$$
Since
beginalign*
lim_xto +infty frac log (x^2lambda) x &= lim_x to+ infty frac 2lambda x^2lambda -1/x^2lambda 1 [mathsf L'hopital rules] \
&= lim_xto +infty frac 2lambda x \
&= 0,
endalign*
we deduce that
$$
lim_n left| log (cos (n^lambda)) right|^1/n = 1.
$$
Therefore the power series converges when $|y| <1$, diverges when $|y|>1$.
Now return to the original question. Using the convergent radius, we know that the series converges when $|8lambda^3| < 1$, i.e. $boldsymbol lambda in (-mathbf 1/mathbf 2, mathbf 0]$; diverges when $boldsymbol lambda < mathbf-1/2$. When $lambda = -1/2$, the series becomes
$$
sum (-1)^n log (cos(n^-1/2)) =colon sum (-1)^n-1 a_n
$$
which is a Leibniz series:
$$
frac 1sqrt n > frac 1sqrt n+1 implies a_n > a_n+1 searrow 0,
$$
hence it converges.
Conclusion: the series diverges when $boldsymbol lambda < -mathbf1/2$, converges when $boldsymbol lambda in [mathbf-1/2, mathbf 0]$. $blacktriangleright$
answered 13 hours ago
xbh
8706
This is very useful and complete, but we have never seen the convergent radius and the power series. I think it will be the first topic of the next course. Anyways, thank you for your answer!
– F.inc
12 hours ago
1
@F.inc Or you could understand this as a practice of Cauchy's root test. Take the $n$th root for the whole term, then you could also deduce the same conclusion. Essentially [to this question], these methods agree.
– xbh
12 hours ago
1
@F.inc If you want to use Dirichlet test, your attempt is capable to use it to show the convergence when $0 >lambda > -1/2$, since the log term monotonically tends to 0, and the rest part forms a convergent series, hence the partial sum is bounded.
– xbh
12 hours ago
1
@F.inc However for $lambda < -1/2$, you would kind of be forced back to use root test to prove the divergence.
– xbh
12 hours ago
add a comment |Â
up vote
2
down vote
I am not sure how to handle $b_n=lnleft(cos(n^lambda)right)$ which tends to $0$ for $ntoinfty$.
Giving $k=-lambdagt0$, can I say that $b_n= Obigg(frac1n^kbigg)$ for $n to infty$?
You rather have $displaystyle b_n= Obigg(frac1n^2kbigg)$. One may use Taylor series expansions to see this, noticing that, as $n to infty$,
$$
cos left(frac1n^k right)=1-frac12n^2k+obigg(frac1n^2kbigg)
$$$$
ln left(cos left(frac1n^k right)right)=lnleft(1-frac12n^2k+obigg(frac1n^2kbigg)right)=-frac12n^2k+obigg(frac1n^2kbigg)=Obigg(frac1n^2kbigg).
$$
answered 14 hours ago
Olivier Oloa
106k17173292
2
Note that the last step in this chain is actually (partly) counterproductive for this problem - see my comment on the original post. On part of the interval, the estimate needs to be bounded away from zero here, not towards it.
– Steven Stadnicki
14 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Solution. $blacktriangleleft$ First determine the radius of convergence for
$$
sum y^n log(cos (n^lambda)).
$$
By Cauchy-Hadamard formula, we need to calculate the following:
beginalign*
left| log (cos (n^lambda)) right|^1/n &= exp left( frac 1n log (- log (cos (n^lambda)))right) \
&= exp left( frac 1n log (-log (1 - n^2lambda/2 + o(n^2lambda)))right)\
&= exp left( frac 1n log (n^2lambda + o(n^2lambda))right).
endalign*
For sufficiently large $n$ we can assume that
$$
frac 12 n ^2lambda leqslant n^2lambda + o(n^2lambda) leqslant frac 32 n^2lambda.
$$
Since
beginalign*
lim_xto +infty frac log (x^2lambda) x &= lim_x to+ infty frac 2lambda x^2lambda -1/x^2lambda 1 [mathsf L'hopital rules] \
&= lim_xto +infty frac 2lambda x \
&= 0,
endalign*
we deduce that
$$
lim_n left| log (cos (n^lambda)) right|^1/n = 1.
$$
Therefore the power series converges when $|y| <1$, diverges when $|y|>1$.
Now return to the original question. Using the convergent radius, we know that the series converges when $|8lambda^3| < 1$, i.e. $boldsymbol lambda in (-mathbf 1/mathbf 2, mathbf 0]$; diverges when $boldsymbol lambda < mathbf-1/2$. When $lambda = -1/2$, the series becomes
$$
sum (-1)^n log (cos(n^-1/2)) =colon sum (-1)^n-1 a_n
$$
which is a Leibniz series:
$$
frac 1sqrt n > frac 1sqrt n+1 implies a_n > a_n+1 searrow 0,
$$
hence it converges.
Conclusion: the series diverges when $boldsymbol lambda < -mathbf1/2$, converges when $boldsymbol lambda in [mathbf-1/2, mathbf 0]$. $blacktriangleright$
answered 13 hours ago
xbh
8706
This is very useful and complete, but we have never seen the convergent radius and the power series. I think it will be the first topic of the next course. Anyways, thank you for your answer!
– F.inc
12 hours ago
1
@F.inc Or you could understand this as a practice of Cauchy's root test. Take the $n$th root for the whole term, then you could also deduce the same conclusion. Essentially [to this question], these methods agree.
– xbh
12 hours ago
1
@F.inc If you want to use Dirichlet test, your attempt is capable to use it to show the convergence when $0 >lambda > -1/2$, since the log term monotonically tends to 0, and the rest part forms a convergent series, hence the partial sum is bounded.
– xbh
12 hours ago
1
@F.inc However for $lambda < -1/2$, you would kind of be forced back to use root test to prove the divergence.
– xbh
12 hours ago
add a comment |Â
up vote
1
down vote
accepted
Solution. $blacktriangleleft$ First determine the radius of convergence for
$$
sum y^n log(cos (n^lambda)).
$$
By Cauchy-Hadamard formula, we need to calculate the following:
beginalign*
left| log (cos (n^lambda)) right|^1/n &= exp left( frac 1n log (- log (cos (n^lambda)))right) \
&= exp left( frac 1n log (-log (1 - n^2lambda/2 + o(n^2lambda)))right)\
&= exp left( frac 1n log (n^2lambda + o(n^2lambda))right).
endalign*
For sufficiently large $n$ we can assume that
$$
frac 12 n ^2lambda leqslant n^2lambda + o(n^2lambda) leqslant frac 32 n^2lambda.
$$
Since
beginalign*
lim_xto +infty frac log (x^2lambda) x &= lim_x to+ infty frac 2lambda x^2lambda -1/x^2lambda 1 [mathsf L'hopital rules] \
&= lim_xto +infty frac 2lambda x \
&= 0,
endalign*
we deduce that
$$
lim_n left| log (cos (n^lambda)) right|^1/n = 1.
$$
Therefore the power series converges when $|y| <1$, diverges when $|y|>1$.
Now return to the original question. Using the convergent radius, we know that the series converges when $|8lambda^3| < 1$, i.e. $boldsymbol lambda in (-mathbf 1/mathbf 2, mathbf 0]$; diverges when $boldsymbol lambda < mathbf-1/2$. When $lambda = -1/2$, the series becomes
$$
sum (-1)^n log (cos(n^-1/2)) =colon sum (-1)^n-1 a_n
$$
which is a Leibniz series:
$$
frac 1sqrt n > frac 1sqrt n+1 implies a_n > a_n+1 searrow 0,
$$
hence it converges.
Conclusion: the series diverges when $boldsymbol lambda < -mathbf1/2$, converges when $boldsymbol lambda in [mathbf-1/2, mathbf 0]$. $blacktriangleright$
answered 13 hours ago
xbh
8706
This is very useful and complete, but we have never seen the convergent radius and the power series. I think it will be the first topic of the next course. Anyways, thank you for your answer!
– F.inc
12 hours ago
1
@F.inc Or you could understand this as a practice of Cauchy's root test. Take the $n$th root for the whole term, then you could also deduce the same conclusion. Essentially [to this question], these methods agree.
– xbh
12 hours ago
1
@F.inc If you want to use Dirichlet test, your attempt is capable to use it to show the convergence when $0 >lambda > -1/2$, since the log term monotonically tends to 0, and the rest part forms a convergent series, hence the partial sum is bounded.
– xbh
12 hours ago
1
@F.inc However for $lambda < -1/2$, you would kind of be forced back to use root test to prove the divergence.
– xbh
12 hours ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Solution. $blacktriangleleft$ First determine the radius of convergence for
$$
sum y^n log(cos (n^lambda)).
$$
By Cauchy-Hadamard formula, we need to calculate the following:
beginalign*
left| log (cos (n^lambda)) right|^1/n &= exp left( frac 1n log (- log (cos (n^lambda)))right) \
&= exp left( frac 1n log (-log (1 - n^2lambda/2 + o(n^2lambda)))right)\
&= exp left( frac 1n log (n^2lambda + o(n^2lambda))right).
endalign*
For sufficiently large $n$ we can assume that
$$
frac 12 n ^2lambda leqslant n^2lambda + o(n^2lambda) leqslant frac 32 n^2lambda.
$$
Since
beginalign*
lim_xto +infty frac log (x^2lambda) x &= lim_x to+ infty frac 2lambda x^2lambda -1/x^2lambda 1 [mathsf L'hopital rules] \
&= lim_xto +infty frac 2lambda x \
&= 0,
endalign*
we deduce that
$$
lim_n left| log (cos (n^lambda)) right|^1/n = 1.
$$
Therefore the power series converges when $|y| <1$, diverges when $|y|>1$.
Now return to the original question. Using the convergent radius, we know that the series converges when $|8lambda^3| < 1$, i.e. $boldsymbol lambda in (-mathbf 1/mathbf 2, mathbf 0]$; diverges when $boldsymbol lambda < mathbf-1/2$. When $lambda = -1/2$, the series becomes
$$
sum (-1)^n log (cos(n^-1/2)) =colon sum (-1)^n-1 a_n
$$
which is a Leibniz series:
$$
frac 1sqrt n > frac 1sqrt n+1 implies a_n > a_n+1 searrow 0,
$$
hence it converges.
Conclusion: the series diverges when $boldsymbol lambda < -mathbf1/2$, converges when $boldsymbol lambda in [mathbf-1/2, mathbf 0]$. $blacktriangleright$
answered 13 hours ago
xbh
8706
Solution. $blacktriangleleft$ First determine the radius of convergence for
$$
sum y^n log(cos (n^lambda)).
$$
By Cauchy-Hadamard formula, we need to calculate the following:
beginalign*
left| log (cos (n^lambda)) right|^1/n &= exp left( frac 1n log (- log (cos (n^lambda)))right) \
&= exp left( frac 1n log (-log (1 - n^2lambda/2 + o(n^2lambda)))right)\
&= exp left( frac 1n log (n^2lambda + o(n^2lambda))right).
endalign*
For sufficiently large $n$ we can assume that
$$
frac 12 n ^2lambda leqslant n^2lambda + o(n^2lambda) leqslant frac 32 n^2lambda.
$$
Since
beginalign*
lim_xto +infty frac log (x^2lambda) x &= lim_x to+ infty frac 2lambda x^2lambda -1/x^2lambda 1 [mathsf L'hopital rules] \
&= lim_xto +infty frac 2lambda x \
&= 0,
endalign*
we deduce that
$$
lim_n left| log (cos (n^lambda)) right|^1/n = 1.
$$
Therefore the power series converges when $|y| <1$, diverges when $|y|>1$.
Now return to the original question. Using the convergent radius, we know that the series converges when $|8lambda^3| < 1$, i.e. $boldsymbol lambda in (-mathbf 1/mathbf 2, mathbf 0]$; diverges when $boldsymbol lambda < mathbf-1/2$. When $lambda = -1/2$, the series becomes
$$
sum (-1)^n log (cos(n^-1/2)) =colon sum (-1)^n-1 a_n
$$
which is a Leibniz series:
$$
frac 1sqrt n > frac 1sqrt n+1 implies a_n > a_n+1 searrow 0,
$$
hence it converges.
Conclusion: the series diverges when $boldsymbol lambda < -mathbf1/2$, converges when $boldsymbol lambda in [mathbf-1/2, mathbf 0]$. $blacktriangleright$
answered 13 hours ago
xbh
8706
answered 13 hours ago
xbh
8706
answered 13 hours ago
xbh
8706
answered 13 hours ago
xbh
8706
8706
This is very useful and complete, but we have never seen the convergent radius and the power series. I think it will be the first topic of the next course. Anyways, thank you for your answer!
– F.inc
12 hours ago
1
@F.inc Or you could understand this as a practice of Cauchy's root test. Take the $n$th root for the whole term, then you could also deduce the same conclusion. Essentially [to this question], these methods agree.
– xbh
12 hours ago
1
@F.inc If you want to use Dirichlet test, your attempt is capable to use it to show the convergence when $0 >lambda > -1/2$, since the log term monotonically tends to 0, and the rest part forms a convergent series, hence the partial sum is bounded.
– xbh
12 hours ago
1
@F.inc However for $lambda < -1/2$, you would kind of be forced back to use root test to prove the divergence.
– xbh
12 hours ago
add a comment |Â
This is very useful and complete, but we have never seen the convergent radius and the power series. I think it will be the first topic of the next course. Anyways, thank you for your answer!
– F.inc
12 hours ago
1
@F.inc Or you could understand this as a practice of Cauchy's root test. Take the $n$th root for the whole term, then you could also deduce the same conclusion. Essentially [to this question], these methods agree.
– xbh
12 hours ago
1
@F.inc If you want to use Dirichlet test, your attempt is capable to use it to show the convergence when $0 >lambda > -1/2$, since the log term monotonically tends to 0, and the rest part forms a convergent series, hence the partial sum is bounded.
– xbh
12 hours ago
1
@F.inc However for $lambda < -1/2$, you would kind of be forced back to use root test to prove the divergence.
– xbh
12 hours ago
This is very useful and complete, but we have never seen the convergent radius and the power series. I think it will be the first topic of the next course. Anyways, thank you for your answer!
– F.inc
12 hours ago
This is very useful and complete, but we have never seen the convergent radius and the power series. I think it will be the first topic of the next course. Anyways, thank you for your answer!
– F.inc
12 hours ago
1
1
@F.inc Or you could understand this as a practice of Cauchy's root test. Take the $n$th root for the whole term, then you could also deduce the same conclusion. Essentially [to this question], these methods agree.
– xbh
12 hours ago
@F.inc Or you could understand this as a practice of Cauchy's root test. Take the $n$th root for the whole term, then you could also deduce the same conclusion. Essentially [to this question], these methods agree.
– xbh
12 hours ago
1
1
@F.inc If you want to use Dirichlet test, your attempt is capable to use it to show the convergence when $0 >lambda > -1/2$, since the log term monotonically tends to 0, and the rest part forms a convergent series, hence the partial sum is bounded.
– xbh
12 hours ago
@F.inc If you want to use Dirichlet test, your attempt is capable to use it to show the convergence when $0 >lambda > -1/2$, since the log term monotonically tends to 0, and the rest part forms a convergent series, hence the partial sum is bounded.
– xbh
12 hours ago
1
1
@F.inc However for $lambda < -1/2$, you would kind of be forced back to use root test to prove the divergence.
– xbh
12 hours ago
@F.inc However for $lambda < -1/2$, you would kind of be forced back to use root test to prove the divergence.
– xbh
12 hours ago
add a comment |Â
up vote
2
down vote
I am not sure how to handle $b_n=lnleft(cos(n^lambda)right)$ which tends to $0$ for $ntoinfty$.
Giving $k=-lambdagt0$, can I say that $b_n= Obigg(frac1n^kbigg)$ for $n to infty$?
You rather have $displaystyle b_n= Obigg(frac1n^2kbigg)$. One may use Taylor series expansions to see this, noticing that, as $n to infty$,
$$
cos left(frac1n^k right)=1-frac12n^2k+obigg(frac1n^2kbigg)
$$$$
ln left(cos left(frac1n^k right)right)=lnleft(1-frac12n^2k+obigg(frac1n^2kbigg)right)=-frac12n^2k+obigg(frac1n^2kbigg)=Obigg(frac1n^2kbigg).
$$
answered 14 hours ago
Olivier Oloa
106k17173292
2
Note that the last step in this chain is actually (partly) counterproductive for this problem - see my comment on the original post. On part of the interval, the estimate needs to be bounded away from zero here, not towards it.
– Steven Stadnicki
14 hours ago
add a comment |Â
up vote
2
down vote
I am not sure how to handle $b_n=lnleft(cos(n^lambda)right)$ which tends to $0$ for $ntoinfty$.
Giving $k=-lambdagt0$, can I say that $b_n= Obigg(frac1n^kbigg)$ for $n to infty$?
You rather have $displaystyle b_n= Obigg(frac1n^2kbigg)$. One may use Taylor series expansions to see this, noticing that, as $n to infty$,
$$
cos left(frac1n^k right)=1-frac12n^2k+obigg(frac1n^2kbigg)
$$$$
ln left(cos left(frac1n^k right)right)=lnleft(1-frac12n^2k+obigg(frac1n^2kbigg)right)=-frac12n^2k+obigg(frac1n^2kbigg)=Obigg(frac1n^2kbigg).
$$
answered 14 hours ago
Olivier Oloa
106k17173292
2
Note that the last step in this chain is actually (partly) counterproductive for this problem - see my comment on the original post. On part of the interval, the estimate needs to be bounded away from zero here, not towards it.
– Steven Stadnicki
14 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I am not sure how to handle $b_n=lnleft(cos(n^lambda)right)$ which tends to $0$ for $ntoinfty$.
Giving $k=-lambdagt0$, can I say that $b_n= Obigg(frac1n^kbigg)$ for $n to infty$?
You rather have $displaystyle b_n= Obigg(frac1n^2kbigg)$. One may use Taylor series expansions to see this, noticing that, as $n to infty$,
$$
cos left(frac1n^k right)=1-frac12n^2k+obigg(frac1n^2kbigg)
$$$$
ln left(cos left(frac1n^k right)right)=lnleft(1-frac12n^2k+obigg(frac1n^2kbigg)right)=-frac12n^2k+obigg(frac1n^2kbigg)=Obigg(frac1n^2kbigg).
$$
answered 14 hours ago
Olivier Oloa
106k17173292
I am not sure how to handle $b_n=lnleft(cos(n^lambda)right)$ which tends to $0$ for $ntoinfty$.
Giving $k=-lambdagt0$, can I say that $b_n= Obigg(frac1n^kbigg)$ for $n to infty$?
You rather have $displaystyle b_n= Obigg(frac1n^2kbigg)$. One may use Taylor series expansions to see this, noticing that, as $n to infty$,
$$
cos left(frac1n^k right)=1-frac12n^2k+obigg(frac1n^2kbigg)
$$$$
ln left(cos left(frac1n^k right)right)=lnleft(1-frac12n^2k+obigg(frac1n^2kbigg)right)=-frac12n^2k+obigg(frac1n^2kbigg)=Obigg(frac1n^2kbigg).
$$
answered 14 hours ago
Olivier Oloa
106k17173292
answered 14 hours ago
Olivier Oloa
106k17173292
answered 14 hours ago
Olivier Oloa
106k17173292
answered 14 hours ago
Olivier Oloa
106k17173292
106k17173292
2
Note that the last step in this chain is actually (partly) counterproductive for this problem - see my comment on the original post. On part of the interval, the estimate needs to be bounded away from zero here, not towards it.
– Steven Stadnicki
14 hours ago
add a comment |Â
2
Note that the last step in this chain is actually (partly) counterproductive for this problem - see my comment on the original post. On part of the interval, the estimate needs to be bounded away from zero here, not towards it.
– Steven Stadnicki
14 hours ago
2
2
Note that the last step in this chain is actually (partly) counterproductive for this problem - see my comment on the original post. On part of the interval, the estimate needs to be bounded away from zero here, not towards it.
– Steven Stadnicki
14 hours ago
Note that the last step in this chain is actually (partly) counterproductive for this problem - see my comment on the original post. On part of the interval, the estimate needs to be bounded away from zero here, not towards it.
– Steven Stadnicki
14 hours ago
add a comment |Â
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2
Saying that $b_n=O(frac1n^lambda)$ - or even $b_n=O(frac1n^2lambda)$, as in the answer below - isn't actually enough in and of itself, because $O()$ doesn't say anything about how small a term can get. Instead, for $lambdagt-1/2$, you need to bound $ln(cos(n^lambda))$ away from zero, to show that it's not so small that it can 'counteract' the growth of $(2lambda)^3n$.
– Steven Stadnicki
14 hours ago
1
@StevenStadnicki My aim was that of using the convergence/divergence of the harmonic series (maybe Leibnitz and Dirichlet when combined with an oscillating one). Do you find that this idea fails? What would you do? Thank you!
– F.inc
13 hours ago
1
The problem with using the divergence of the geometric series for $lambdalt-1/2$ is that you have to show, not that $ln(cos(n^lambda))$ is small enough, but that it's large enough; in principle it's possible that, for instance, $ln(cos(n^lambda))lt n^-n$, in which case the presence of the geometrically-growing factor of $(2lambda)^3n$ wouldn't be enough to guarantee divergence. This doesn't actually happen, but you have to prove that it doesn't happen.
– Steven Stadnicki
11 hours ago
1
The idea of using the geometric series for comparative purposes is a good one, but you have to make sure that your bounds are correct. (Also, you should make sure to handle $lambda=-1/2$ explicitly as well.)
– Steven Stadnicki
11 hours ago