Mixing
Study the convergence of the series $sum_{n=1}^inftyleft(ln(ntanfrac{1}{n})right)^{1/3}$
$begingroup$
Study the convergence of the $$sum_{n=1}^infty ln^{1/3}left(ntanfrac{1}{n}right)$$
I'm trying to study the convergence of the series and I need a little help.
This is what I tried to do.
I used the limit comparison test
$$lim_{nto infty}frac{ln(ntanfrac{1}{n})}{ntanfrac{1}{n}-1}=1$$
This implies that
$$sum_{n=1}^inftyleft(ln(ntanfrac{1}{n})right)^frac{1}{3} sim sum_{n=1}^inftyleft(ntanfrac{1}{n}-1right)^frac{1}{3}$$
And this is where I got stuck because I don't know how to study the convergence of the new one. Should I try another method or should I proceed with this one? If so how can I do that?
The answer in the solution says that it diverges
calculus sequences-and-series
asked Feb 1 at 0:03
J.DaneJ.Dane
382214
$endgroup$
add a comment |
$begingroup$
Study the convergence of the $$sum_{n=1}^infty ln^{1/3}left(ntanfrac{1}{n}right)$$
I'm trying to study the convergence of the series and I need a little help.
This is what I tried to do.
I used the limit comparison test
$$lim_{nto infty}frac{ln(ntanfrac{1}{n})}{ntanfrac{1}{n}-1}=1$$
This implies that
$$sum_{n=1}^inftyleft(ln(ntanfrac{1}{n})right)^frac{1}{3} sim sum_{n=1}^inftyleft(ntanfrac{1}{n}-1right)^frac{1}{3}$$
And this is where I got stuck because I don't know how to study the convergence of the new one. Should I try another method or should I proceed with this one? If so how can I do that?
The answer in the solution says that it diverges
calculus sequences-and-series
asked Feb 1 at 0:03
J.DaneJ.Dane
382214
$endgroup$
$begingroup$
You are almost there: it happens that $f(x)=frac1xtan x-1$ is of order $x^2$ when $xto0$, hence the $n$th term of the series is of order $left(frac1{n^2}right)^{1/3}$ and the series diverges. Now, how you should prove that $f(x)simfrac13x^2$ depends very much on your background. If you can invoke the Taylor expansion of $tan$ at order $3$, then your proof is complete.
$endgroup$
– Did
Feb 1 at 7:16
add a comment |
$begingroup$
Study the convergence of the $$sum_{n=1}^infty ln^{1/3}left(ntanfrac{1}{n}right)$$
I'm trying to study the convergence of the series and I need a little help.
This is what I tried to do.
I used the limit comparison test
$$lim_{nto infty}frac{ln(ntanfrac{1}{n})}{ntanfrac{1}{n}-1}=1$$
This implies that
$$sum_{n=1}^inftyleft(ln(ntanfrac{1}{n})right)^frac{1}{3} sim sum_{n=1}^inftyleft(ntanfrac{1}{n}-1right)^frac{1}{3}$$
And this is where I got stuck because I don't know how to study the convergence of the new one. Should I try another method or should I proceed with this one? If so how can I do that?
The answer in the solution says that it diverges
calculus sequences-and-series
asked Feb 1 at 0:03
J.DaneJ.Dane
382214
$endgroup$
Study the convergence of the $$sum_{n=1}^infty ln^{1/3}left(ntanfrac{1}{n}right)$$
I'm trying to study the convergence of the series and I need a little help.
This is what I tried to do.
I used the limit comparison test
$$lim_{nto infty}frac{ln(ntanfrac{1}{n})}{ntanfrac{1}{n}-1}=1$$
This implies that
$$sum_{n=1}^inftyleft(ln(ntanfrac{1}{n})right)^frac{1}{3} sim sum_{n=1}^inftyleft(ntanfrac{1}{n}-1right)^frac{1}{3}$$
And this is where I got stuck because I don't know how to study the convergence of the new one. Should I try another method or should I proceed with this one? If so how can I do that?
The answer in the solution says that it diverges
calculus sequences-and-series
calculus sequences-and-series
asked Feb 1 at 0:03
J.DaneJ.Dane
382214
asked Feb 1 at 0:03
J.DaneJ.Dane
382214
edited Feb 1 at 0:13
J.Dane
asked Feb 1 at 0:03
J.DaneJ.Dane
382214
asked Feb 1 at 0:03
J.DaneJ.Dane
382214
asked Feb 1 at 0:03
J.DaneJ.Dane
382214
382214
$begingroup$
You are almost there: it happens that $f(x)=frac1xtan x-1$ is of order $x^2$ when $xto0$, hence the $n$th term of the series is of order $left(frac1{n^2}right)^{1/3}$ and the series diverges. Now, how you should prove that $f(x)simfrac13x^2$ depends very much on your background. If you can invoke the Taylor expansion of $tan$ at order $3$, then your proof is complete.
$endgroup$
– Did
Feb 1 at 7:16
add a comment |
$begingroup$
You are almost there: it happens that $f(x)=frac1xtan x-1$ is of order $x^2$ when $xto0$, hence the $n$th term of the series is of order $left(frac1{n^2}right)^{1/3}$ and the series diverges. Now, how you should prove that $f(x)simfrac13x^2$ depends very much on your background. If you can invoke the Taylor expansion of $tan$ at order $3$, then your proof is complete.
$endgroup$
– Did
Feb 1 at 7:16
$begingroup$
You are almost there: it happens that $f(x)=frac1xtan x-1$ is of order $x^2$ when $xto0$, hence the $n$th term of the series is of order $left(frac1{n^2}right)^{1/3}$ and the series diverges. Now, how you should prove that $f(x)simfrac13x^2$ depends very much on your background. If you can invoke the Taylor expansion of $tan$ at order $3$, then your proof is complete.
$endgroup$
– Did
Feb 1 at 7:16
$begingroup$
You are almost there: it happens that $f(x)=frac1xtan x-1$ is of order $x^2$ when $xto0$, hence the $n$th term of the series is of order $left(frac1{n^2}right)^{1/3}$ and the series diverges. Now, how you should prove that $f(x)simfrac13x^2$ depends very much on your background. If you can invoke the Taylor expansion of $tan$ at order $3$, then your proof is complete.
$endgroup$
– Did
Feb 1 at 7:16
add a comment |
1 Answer
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$begingroup$
For $n$ sufficiently large we have $$tan {1over n}={1over n}+{1over 3n^3}+cdots$$ according to MaclaurinSeries therefore $$forall nin Bbb Nquad,quad tan {1over n}>{1over n}+{1over 3n^3}$$which can be also verified from https://www.desmos.com/calculator/2zjhgxlelr. By substitution we obtain $$left(ntan {1over n}-1right)^{1over 3}{>left(1+{1over 3n^2}-1right)^{1over 3}\={1over sqrt[3]3 n^{2over 3}}\>{1over 2n}}$$as $sum {1over 2n}$ diverges, so do $$sum left(ntan {1over n}-1right)^{1over 3}$$and $$sum_{n=1}^infty ln^{1/3}left(ntanfrac{1}{n}right)$$
answered Feb 2 at 15:01
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
I understand what you tried to do but I don't think you that's the Maclaurin series for $tanfrac{1}{n}$ shouldn't it be $tanfrac{1}{n}=frac{1}{n}+frac{1}{3n^3}+frac{2}{15n^5}+...$
$endgroup$
– J.Dane
Feb 2 at 18:09
$begingroup$
Thank you for the feedback!!
$endgroup$
– Mostafa Ayaz
Feb 2 at 22:45
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For $n$ sufficiently large we have $$tan {1over n}={1over n}+{1over 3n^3}+cdots$$ according to MaclaurinSeries therefore $$forall nin Bbb Nquad,quad tan {1over n}>{1over n}+{1over 3n^3}$$which can be also verified from https://www.desmos.com/calculator/2zjhgxlelr. By substitution we obtain $$left(ntan {1over n}-1right)^{1over 3}{>left(1+{1over 3n^2}-1right)^{1over 3}\={1over sqrt[3]3 n^{2over 3}}\>{1over 2n}}$$as $sum {1over 2n}$ diverges, so do $$sum left(ntan {1over n}-1right)^{1over 3}$$and $$sum_{n=1}^infty ln^{1/3}left(ntanfrac{1}{n}right)$$
answered Feb 2 at 15:01
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
I understand what you tried to do but I don't think you that's the Maclaurin series for $tanfrac{1}{n}$ shouldn't it be $tanfrac{1}{n}=frac{1}{n}+frac{1}{3n^3}+frac{2}{15n^5}+...$
$endgroup$
– J.Dane
Feb 2 at 18:09
$begingroup$
Thank you for the feedback!!
$endgroup$
– Mostafa Ayaz
Feb 2 at 22:45
add a comment |
$begingroup$
For $n$ sufficiently large we have $$tan {1over n}={1over n}+{1over 3n^3}+cdots$$ according to MaclaurinSeries therefore $$forall nin Bbb Nquad,quad tan {1over n}>{1over n}+{1over 3n^3}$$which can be also verified from https://www.desmos.com/calculator/2zjhgxlelr. By substitution we obtain $$left(ntan {1over n}-1right)^{1over 3}{>left(1+{1over 3n^2}-1right)^{1over 3}\={1over sqrt[3]3 n^{2over 3}}\>{1over 2n}}$$as $sum {1over 2n}$ diverges, so do $$sum left(ntan {1over n}-1right)^{1over 3}$$and $$sum_{n=1}^infty ln^{1/3}left(ntanfrac{1}{n}right)$$
answered Feb 2 at 15:01
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
I understand what you tried to do but I don't think you that's the Maclaurin series for $tanfrac{1}{n}$ shouldn't it be $tanfrac{1}{n}=frac{1}{n}+frac{1}{3n^3}+frac{2}{15n^5}+...$
$endgroup$
– J.Dane
Feb 2 at 18:09
$begingroup$
Thank you for the feedback!!
$endgroup$
– Mostafa Ayaz
Feb 2 at 22:45
add a comment |
$begingroup$
For $n$ sufficiently large we have $$tan {1over n}={1over n}+{1over 3n^3}+cdots$$ according to MaclaurinSeries therefore $$forall nin Bbb Nquad,quad tan {1over n}>{1over n}+{1over 3n^3}$$which can be also verified from https://www.desmos.com/calculator/2zjhgxlelr. By substitution we obtain $$left(ntan {1over n}-1right)^{1over 3}{>left(1+{1over 3n^2}-1right)^{1over 3}\={1over sqrt[3]3 n^{2over 3}}\>{1over 2n}}$$as $sum {1over 2n}$ diverges, so do $$sum left(ntan {1over n}-1right)^{1over 3}$$and $$sum_{n=1}^infty ln^{1/3}left(ntanfrac{1}{n}right)$$
answered Feb 2 at 15:01
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
For $n$ sufficiently large we have $$tan {1over n}={1over n}+{1over 3n^3}+cdots$$ according to MaclaurinSeries therefore $$forall nin Bbb Nquad,quad tan {1over n}>{1over n}+{1over 3n^3}$$which can be also verified from https://www.desmos.com/calculator/2zjhgxlelr. By substitution we obtain $$left(ntan {1over n}-1right)^{1over 3}{>left(1+{1over 3n^2}-1right)^{1over 3}\={1over sqrt[3]3 n^{2over 3}}\>{1over 2n}}$$as $sum {1over 2n}$ diverges, so do $$sum left(ntan {1over n}-1right)^{1over 3}$$and $$sum_{n=1}^infty ln^{1/3}left(ntanfrac{1}{n}right)$$
answered Feb 2 at 15:01
Mostafa AyazMostafa Ayaz
18.1k31040
edited Feb 2 at 22:45
answered Feb 2 at 15:01
Mostafa AyazMostafa Ayaz
18.1k31040
answered Feb 2 at 15:01
Mostafa AyazMostafa Ayaz
18.1k31040
answered Feb 2 at 15:01
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
$begingroup$
I understand what you tried to do but I don't think you that's the Maclaurin series for $tanfrac{1}{n}$ shouldn't it be $tanfrac{1}{n}=frac{1}{n}+frac{1}{3n^3}+frac{2}{15n^5}+...$
$endgroup$
– J.Dane
Feb 2 at 18:09
$begingroup$
Thank you for the feedback!!
$endgroup$
– Mostafa Ayaz
Feb 2 at 22:45
add a comment |
$begingroup$
I understand what you tried to do but I don't think you that's the Maclaurin series for $tanfrac{1}{n}$ shouldn't it be $tanfrac{1}{n}=frac{1}{n}+frac{1}{3n^3}+frac{2}{15n^5}+...$
$endgroup$
– J.Dane
Feb 2 at 18:09
$begingroup$
Thank you for the feedback!!
$endgroup$
– Mostafa Ayaz
Feb 2 at 22:45
$begingroup$
I understand what you tried to do but I don't think you that's the Maclaurin series for $tanfrac{1}{n}$ shouldn't it be $tanfrac{1}{n}=frac{1}{n}+frac{1}{3n^3}+frac{2}{15n^5}+...$
$endgroup$
– J.Dane
Feb 2 at 18:09
$begingroup$
I understand what you tried to do but I don't think you that's the Maclaurin series for $tanfrac{1}{n}$ shouldn't it be $tanfrac{1}{n}=frac{1}{n}+frac{1}{3n^3}+frac{2}{15n^5}+...$
$endgroup$
– J.Dane
Feb 2 at 18:09
$begingroup$
Thank you for the feedback!!
$endgroup$
– Mostafa Ayaz
Feb 2 at 22:45
$begingroup$
Thank you for the feedback!!
$endgroup$
– Mostafa Ayaz
Feb 2 at 22:45
add a comment |
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$begingroup$
You are almost there: it happens that $f(x)=frac1xtan x-1$ is of order $x^2$ when $xto0$, hence the $n$th term of the series is of order $left(frac1{n^2}right)^{1/3}$ and the series diverges. Now, how you should prove that $f(x)simfrac13x^2$ depends very much on your background. If you can invoke the Taylor expansion of $tan$ at order $3$, then your proof is complete.
$endgroup$
– Did
Feb 1 at 7:16