Mixing
Determine for what values of $x$ the series converges.
$begingroup$
Determine for what values of $x$ the series converges.
Here is the series:
$$sum_{n = 1}^{infty} frac{x^n}{sin^n n}.$$
My trial:
Considering $a_{n} = frac{1}{sin^n n} $
I calculated $|frac{a_{n+1} x^{n+1}}{a_n x^n}|$ which is equal to $|x||frac{sin^n n}{sin^{n+1}(n+1)}|$ then I want to take the limit as $n rightarrow infty$ but then what shall I do? I do not know how to solve this limit
real-analysis calculus sequences-and-series analysis
asked Jan 29 at 5:27
hopefullyhopefully
304214
$endgroup$
|
show 4 more comments
$begingroup$
Determine for what values of $x$ the series converges.
Here is the series:
$$sum_{n = 1}^{infty} frac{x^n}{sin^n n}.$$
My trial:
Considering $a_{n} = frac{1}{sin^n n} $
I calculated $|frac{a_{n+1} x^{n+1}}{a_n x^n}|$ which is equal to $|x||frac{sin^n n}{sin^{n+1}(n+1)}|$ then I want to take the limit as $n rightarrow infty$ but then what shall I do? I do not know how to solve this limit
real-analysis calculus sequences-and-series analysis
asked Jan 29 at 5:27
hopefullyhopefully
304214
$endgroup$
$begingroup$
you have a power series, and can simply compute its radius of convergence en.wikipedia.org/wiki/Radius_of_convergence
$endgroup$
– Hayk
Jan 29 at 6:05
$begingroup$
I know this information and this is what I am trying to do but I have a difficulty which I mentioned above.@Hayk
$endgroup$
– hopefully
Jan 29 at 6:08
2
$begingroup$
Have you tried the root test, along with the fact that $sin$ is dense in $[-1,1]$?
$endgroup$
– Clement C.
Jan 29 at 6:15
$begingroup$
No I will try it @ClementC.
$endgroup$
– hopefully
Jan 29 at 6:18
$begingroup$
@ClementC. Ithink the limit will not exist by the root test because I will calculate $lim (1/sin n)$.... it seems like the root test also does not work.
$endgroup$
– hopefully
Jan 29 at 6:23
|
show 4 more comments
$begingroup$
Determine for what values of $x$ the series converges.
Here is the series:
$$sum_{n = 1}^{infty} frac{x^n}{sin^n n}.$$
My trial:
Considering $a_{n} = frac{1}{sin^n n} $
I calculated $|frac{a_{n+1} x^{n+1}}{a_n x^n}|$ which is equal to $|x||frac{sin^n n}{sin^{n+1}(n+1)}|$ then I want to take the limit as $n rightarrow infty$ but then what shall I do? I do not know how to solve this limit
real-analysis calculus sequences-and-series analysis
asked Jan 29 at 5:27
hopefullyhopefully
304214
$endgroup$
Determine for what values of $x$ the series converges.
Here is the series:
$$sum_{n = 1}^{infty} frac{x^n}{sin^n n}.$$
My trial:
Considering $a_{n} = frac{1}{sin^n n} $
I calculated $|frac{a_{n+1} x^{n+1}}{a_n x^n}|$ which is equal to $|x||frac{sin^n n}{sin^{n+1}(n+1)}|$ then I want to take the limit as $n rightarrow infty$ but then what shall I do? I do not know how to solve this limit
real-analysis calculus sequences-and-series analysis
real-analysis calculus sequences-and-series analysis
asked Jan 29 at 5:27
hopefullyhopefully
304214
asked Jan 29 at 5:27
hopefullyhopefully
304214
edited Jan 29 at 5:49
hopefully
asked Jan 29 at 5:27
hopefullyhopefully
304214
asked Jan 29 at 5:27
hopefullyhopefully
304214
asked Jan 29 at 5:27
hopefullyhopefully
304214
304214
$begingroup$
you have a power series, and can simply compute its radius of convergence en.wikipedia.org/wiki/Radius_of_convergence
$endgroup$
– Hayk
Jan 29 at 6:05
$begingroup$
I know this information and this is what I am trying to do but I have a difficulty which I mentioned above.@Hayk
$endgroup$
– hopefully
Jan 29 at 6:08
2
$begingroup$
Have you tried the root test, along with the fact that $sin$ is dense in $[-1,1]$?
$endgroup$
– Clement C.
Jan 29 at 6:15
$begingroup$
No I will try it @ClementC.
$endgroup$
– hopefully
Jan 29 at 6:18
$begingroup$
@ClementC. Ithink the limit will not exist by the root test because I will calculate $lim (1/sin n)$.... it seems like the root test also does not work.
$endgroup$
– hopefully
Jan 29 at 6:23
|
show 4 more comments
$begingroup$
you have a power series, and can simply compute its radius of convergence en.wikipedia.org/wiki/Radius_of_convergence
$endgroup$
– Hayk
Jan 29 at 6:05
$begingroup$
I know this information and this is what I am trying to do but I have a difficulty which I mentioned above.@Hayk
$endgroup$
– hopefully
Jan 29 at 6:08
2
$begingroup$
Have you tried the root test, along with the fact that $sin$ is dense in $[-1,1]$?
$endgroup$
– Clement C.
Jan 29 at 6:15
$begingroup$
No I will try it @ClementC.
$endgroup$
– hopefully
Jan 29 at 6:18
$begingroup$
@ClementC. Ithink the limit will not exist by the root test because I will calculate $lim (1/sin n)$.... it seems like the root test also does not work.
$endgroup$
– hopefully
Jan 29 at 6:23
$begingroup$
you have a power series, and can simply compute its radius of convergence en.wikipedia.org/wiki/Radius_of_convergence
$endgroup$
– Hayk
Jan 29 at 6:05
$begingroup$
you have a power series, and can simply compute its radius of convergence en.wikipedia.org/wiki/Radius_of_convergence
$endgroup$
– Hayk
Jan 29 at 6:05
$begingroup$
I know this information and this is what I am trying to do but I have a difficulty which I mentioned above.@Hayk
$endgroup$
– hopefully
Jan 29 at 6:08
$begingroup$
I know this information and this is what I am trying to do but I have a difficulty which I mentioned above.@Hayk
$endgroup$
– hopefully
Jan 29 at 6:08
2
2
$begingroup$
Have you tried the root test, along with the fact that $sin$ is dense in $[-1,1]$?
$endgroup$
– Clement C.
Jan 29 at 6:15
$begingroup$
Have you tried the root test, along with the fact that $sin$ is dense in $[-1,1]$?
$endgroup$
– Clement C.
Jan 29 at 6:15
$begingroup$
No I will try it @ClementC.
$endgroup$
– hopefully
Jan 29 at 6:18
$begingroup$
No I will try it @ClementC.
$endgroup$
– hopefully
Jan 29 at 6:18
$begingroup$
@ClementC. Ithink the limit will not exist by the root test because I will calculate $lim (1/sin n)$.... it seems like the root test also does not work.
$endgroup$
– hopefully
Jan 29 at 6:23
$begingroup$
@ClementC. Ithink the limit will not exist by the root test because I will calculate $lim (1/sin n)$.... it seems like the root test also does not work.
$endgroup$
– hopefully
Jan 29 at 6:23
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
(Big) hint: use the root test, along with the fact that $(sin n)_n$ is dense in $[-1,1]$.
In more detail (put your mouse on the area to reveal it):
Since $(sin n)_n$ is dense in $[-1,1]$, there exists a sequence of integers $(n_k)_k$ such that $sin n_k xrightarrow[ktoinfty]{} 0^+$, from which $sqrt[n_k]{a_{n_k}} = frac{1}{sin n_k} xrightarrow[ktoinfty]{} +infty$. This means that $limsup_n sqrt[n]{a_n} = +infty$: applying the root test (with $Cstackrel{rm def}{=}infty$), you therefore get that the radius of convergence is $frac{1}{limsup_n a_n} = 0$.
answered Jan 29 at 6:19
Clement C.Clement C.
51k34093
$endgroup$
$begingroup$
could you write the detailed steps for me as I am confused about |sin n| how I will write its limit?
$endgroup$
– hopefully
Jan 29 at 7:26
$begingroup$
@hopefully Done, see my edit. Once again, the root test asks you to consider the $limsup$ of $sqrt[n]{a_n}$ (which exists), not the limit (which may not exist, and here indeed does not exist).
$endgroup$
– Clement C.
Jan 29 at 7:33
$begingroup$
I need a proof that the limtsup of sin n is +infinity
$endgroup$
– hopefully
Jan 29 at 7:35
1
$begingroup$
It's in the answer (in the "spoiler"). It's not of $sin n$, it's of $sqrt[n]{a_n}$.
$endgroup$
– Clement C.
Jan 29 at 7:37
1
$begingroup$
Yes. The largest. But we want the limsup of 1/sin n, NOT of sin n.
$endgroup$
– Clement C.
Jan 29 at 7:46
|
show 14 more comments
Your Answer
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1 Answer
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1 Answer
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$begingroup$
(Big) hint: use the root test, along with the fact that $(sin n)_n$ is dense in $[-1,1]$.
In more detail (put your mouse on the area to reveal it):
Since $(sin n)_n$ is dense in $[-1,1]$, there exists a sequence of integers $(n_k)_k$ such that $sin n_k xrightarrow[ktoinfty]{} 0^+$, from which $sqrt[n_k]{a_{n_k}} = frac{1}{sin n_k} xrightarrow[ktoinfty]{} +infty$. This means that $limsup_n sqrt[n]{a_n} = +infty$: applying the root test (with $Cstackrel{rm def}{=}infty$), you therefore get that the radius of convergence is $frac{1}{limsup_n a_n} = 0$.
answered Jan 29 at 6:19
Clement C.Clement C.
51k34093
$endgroup$
$begingroup$
could you write the detailed steps for me as I am confused about |sin n| how I will write its limit?
$endgroup$
– hopefully
Jan 29 at 7:26
$begingroup$
@hopefully Done, see my edit. Once again, the root test asks you to consider the $limsup$ of $sqrt[n]{a_n}$ (which exists), not the limit (which may not exist, and here indeed does not exist).
$endgroup$
– Clement C.
Jan 29 at 7:33
$begingroup$
I need a proof that the limtsup of sin n is +infinity
$endgroup$
– hopefully
Jan 29 at 7:35
1
$begingroup$
It's in the answer (in the "spoiler"). It's not of $sin n$, it's of $sqrt[n]{a_n}$.
$endgroup$
– Clement C.
Jan 29 at 7:37
1
$begingroup$
Yes. The largest. But we want the limsup of 1/sin n, NOT of sin n.
$endgroup$
– Clement C.
Jan 29 at 7:46
|
show 14 more comments
$begingroup$
(Big) hint: use the root test, along with the fact that $(sin n)_n$ is dense in $[-1,1]$.
In more detail (put your mouse on the area to reveal it):
Since $(sin n)_n$ is dense in $[-1,1]$, there exists a sequence of integers $(n_k)_k$ such that $sin n_k xrightarrow[ktoinfty]{} 0^+$, from which $sqrt[n_k]{a_{n_k}} = frac{1}{sin n_k} xrightarrow[ktoinfty]{} +infty$. This means that $limsup_n sqrt[n]{a_n} = +infty$: applying the root test (with $Cstackrel{rm def}{=}infty$), you therefore get that the radius of convergence is $frac{1}{limsup_n a_n} = 0$.
answered Jan 29 at 6:19
Clement C.Clement C.
51k34093
$endgroup$
$begingroup$
could you write the detailed steps for me as I am confused about |sin n| how I will write its limit?
$endgroup$
– hopefully
Jan 29 at 7:26
$begingroup$
@hopefully Done, see my edit. Once again, the root test asks you to consider the $limsup$ of $sqrt[n]{a_n}$ (which exists), not the limit (which may not exist, and here indeed does not exist).
$endgroup$
– Clement C.
Jan 29 at 7:33
$begingroup$
I need a proof that the limtsup of sin n is +infinity
$endgroup$
– hopefully
Jan 29 at 7:35
1
$begingroup$
It's in the answer (in the "spoiler"). It's not of $sin n$, it's of $sqrt[n]{a_n}$.
$endgroup$
– Clement C.
Jan 29 at 7:37
1
$begingroup$
Yes. The largest. But we want the limsup of 1/sin n, NOT of sin n.
$endgroup$
– Clement C.
Jan 29 at 7:46
|
show 14 more comments
$begingroup$
(Big) hint: use the root test, along with the fact that $(sin n)_n$ is dense in $[-1,1]$.
In more detail (put your mouse on the area to reveal it):
Since $(sin n)_n$ is dense in $[-1,1]$, there exists a sequence of integers $(n_k)_k$ such that $sin n_k xrightarrow[ktoinfty]{} 0^+$, from which $sqrt[n_k]{a_{n_k}} = frac{1}{sin n_k} xrightarrow[ktoinfty]{} +infty$. This means that $limsup_n sqrt[n]{a_n} = +infty$: applying the root test (with $Cstackrel{rm def}{=}infty$), you therefore get that the radius of convergence is $frac{1}{limsup_n a_n} = 0$.
answered Jan 29 at 6:19
Clement C.Clement C.
51k34093
$endgroup$
(Big) hint: use the root test, along with the fact that $(sin n)_n$ is dense in $[-1,1]$.
In more detail (put your mouse on the area to reveal it):
Since $(sin n)_n$ is dense in $[-1,1]$, there exists a sequence of integers $(n_k)_k$ such that $sin n_k xrightarrow[ktoinfty]{} 0^+$, from which $sqrt[n_k]{a_{n_k}} = frac{1}{sin n_k} xrightarrow[ktoinfty]{} +infty$. This means that $limsup_n sqrt[n]{a_n} = +infty$: applying the root test (with $Cstackrel{rm def}{=}infty$), you therefore get that the radius of convergence is $frac{1}{limsup_n a_n} = 0$.
answered Jan 29 at 6:19
Clement C.Clement C.
51k34093
edited Jan 29 at 7:32
answered Jan 29 at 6:19
Clement C.Clement C.
51k34093
answered Jan 29 at 6:19
Clement C.Clement C.
51k34093
answered Jan 29 at 6:19
Clement C.Clement C.
51k34093
51k34093
$begingroup$
could you write the detailed steps for me as I am confused about |sin n| how I will write its limit?
$endgroup$
– hopefully
Jan 29 at 7:26
$begingroup$
@hopefully Done, see my edit. Once again, the root test asks you to consider the $limsup$ of $sqrt[n]{a_n}$ (which exists), not the limit (which may not exist, and here indeed does not exist).
$endgroup$
– Clement C.
Jan 29 at 7:33
$begingroup$
I need a proof that the limtsup of sin n is +infinity
$endgroup$
– hopefully
Jan 29 at 7:35
1
$begingroup$
It's in the answer (in the "spoiler"). It's not of $sin n$, it's of $sqrt[n]{a_n}$.
$endgroup$
– Clement C.
Jan 29 at 7:37
1
$begingroup$
Yes. The largest. But we want the limsup of 1/sin n, NOT of sin n.
$endgroup$
– Clement C.
Jan 29 at 7:46
|
show 14 more comments
$begingroup$
could you write the detailed steps for me as I am confused about |sin n| how I will write its limit?
$endgroup$
– hopefully
Jan 29 at 7:26
$begingroup$
@hopefully Done, see my edit. Once again, the root test asks you to consider the $limsup$ of $sqrt[n]{a_n}$ (which exists), not the limit (which may not exist, and here indeed does not exist).
$endgroup$
– Clement C.
Jan 29 at 7:33
$begingroup$
I need a proof that the limtsup of sin n is +infinity
$endgroup$
– hopefully
Jan 29 at 7:35
1
$begingroup$
It's in the answer (in the "spoiler"). It's not of $sin n$, it's of $sqrt[n]{a_n}$.
$endgroup$
– Clement C.
Jan 29 at 7:37
1
$begingroup$
Yes. The largest. But we want the limsup of 1/sin n, NOT of sin n.
$endgroup$
– Clement C.
Jan 29 at 7:46
$begingroup$
could you write the detailed steps for me as I am confused about |sin n| how I will write its limit?
$endgroup$
– hopefully
Jan 29 at 7:26
$begingroup$
could you write the detailed steps for me as I am confused about |sin n| how I will write its limit?
$endgroup$
– hopefully
Jan 29 at 7:26
$begingroup$
@hopefully Done, see my edit. Once again, the root test asks you to consider the $limsup$ of $sqrt[n]{a_n}$ (which exists), not the limit (which may not exist, and here indeed does not exist).
$endgroup$
– Clement C.
Jan 29 at 7:33
$begingroup$
@hopefully Done, see my edit. Once again, the root test asks you to consider the $limsup$ of $sqrt[n]{a_n}$ (which exists), not the limit (which may not exist, and here indeed does not exist).
$endgroup$
– Clement C.
Jan 29 at 7:33
$begingroup$
I need a proof that the limtsup of sin n is +infinity
$endgroup$
– hopefully
Jan 29 at 7:35
$begingroup$
I need a proof that the limtsup of sin n is +infinity
$endgroup$
– hopefully
Jan 29 at 7:35
1
1
$begingroup$
It's in the answer (in the "spoiler"). It's not of $sin n$, it's of $sqrt[n]{a_n}$.
$endgroup$
– Clement C.
Jan 29 at 7:37
$begingroup$
It's in the answer (in the "spoiler"). It's not of $sin n$, it's of $sqrt[n]{a_n}$.
$endgroup$
– Clement C.
Jan 29 at 7:37
1
1
$begingroup$
Yes. The largest. But we want the limsup of 1/sin n, NOT of sin n.
$endgroup$
– Clement C.
Jan 29 at 7:46
$begingroup$
Yes. The largest. But we want the limsup of 1/sin n, NOT of sin n.
$endgroup$
– Clement C.
Jan 29 at 7:46
|
show 14 more comments
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$begingroup$
you have a power series, and can simply compute its radius of convergence en.wikipedia.org/wiki/Radius_of_convergence
$endgroup$
– Hayk
Jan 29 at 6:05
$begingroup$
I know this information and this is what I am trying to do but I have a difficulty which I mentioned above.@Hayk
$endgroup$
– hopefully
Jan 29 at 6:08
2
$begingroup$
Have you tried the root test, along with the fact that $sin$ is dense in $[-1,1]$?
$endgroup$
– Clement C.
Jan 29 at 6:15
$begingroup$
No I will try it @ClementC.
$endgroup$
– hopefully
Jan 29 at 6:18
$begingroup$
@ClementC. Ithink the limit will not exist by the root test because I will calculate $lim (1/sin n)$.... it seems like the root test also does not work.
$endgroup$
– hopefully
Jan 29 at 6:23