$x^n+nx-1$ has a unique solution












7












$begingroup$


Show that for any integer $ngeq1,$ the equation $$x^n+nx-1=0$$ has a unique positive solution $x_n$. Furthermore, show that $x_n$ is such that for any $p>1$ the series $sum_{n=1}^{infty}x_n^p$ is convergent.



For the first part of the question I can prove the solution by the intermediate value theorem (by considering $x=0$ and $x=1$). And also uniqueness is achieved because the function is increasing (since the first derivative is always positive.)



But how about the convergence of the series?






real-analysis calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    You mention a $P > 1$ but it is not present anywhere in the series. From the wording, I would expect the sum of $x_n^P$
    $endgroup$
    – Will Jagy
    Dec 19 '18 at 22:30












  • $begingroup$
    Sorry.I edited it now
    $endgroup$
    – DD90
    Dec 19 '18 at 22:31
















7












$begingroup$


Show that for any integer $ngeq1,$ the equation $$x^n+nx-1=0$$ has a unique positive solution $x_n$. Furthermore, show that $x_n$ is such that for any $p>1$ the series $sum_{n=1}^{infty}x_n^p$ is convergent.



For the first part of the question I can prove the solution by the intermediate value theorem (by considering $x=0$ and $x=1$). And also uniqueness is achieved because the function is increasing (since the first derivative is always positive.)



But how about the convergence of the series?






real-analysis calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    You mention a $P > 1$ but it is not present anywhere in the series. From the wording, I would expect the sum of $x_n^P$
    $endgroup$
    – Will Jagy
    Dec 19 '18 at 22:30












  • $begingroup$
    Sorry.I edited it now
    $endgroup$
    – DD90
    Dec 19 '18 at 22:31














7












7








7


2



$begingroup$


Show that for any integer $ngeq1,$ the equation $$x^n+nx-1=0$$ has a unique positive solution $x_n$. Furthermore, show that $x_n$ is such that for any $p>1$ the series $sum_{n=1}^{infty}x_n^p$ is convergent.



For the first part of the question I can prove the solution by the intermediate value theorem (by considering $x=0$ and $x=1$). And also uniqueness is achieved because the function is increasing (since the first derivative is always positive.)



But how about the convergence of the series?






real-analysis calculus







share|cite|improve this question











$endgroup$




Show that for any integer $ngeq1,$ the equation $$x^n+nx-1=0$$ has a unique positive solution $x_n$. Furthermore, show that $x_n$ is such that for any $p>1$ the series $sum_{n=1}^{infty}x_n^p$ is convergent.



For the first part of the question I can prove the solution by the intermediate value theorem (by considering $x=0$ and $x=1$). And also uniqueness is achieved because the function is increasing (since the first derivative is always positive.)



But how about the convergence of the series?






real-analysis calculus




real-analysis calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 22:34









mathworker21

9,0061928




9,0061928










asked Dec 19 '18 at 22:27









DD90DD90

2648




2648












  • $begingroup$
    You mention a $P > 1$ but it is not present anywhere in the series. From the wording, I would expect the sum of $x_n^P$
    $endgroup$
    – Will Jagy
    Dec 19 '18 at 22:30












  • $begingroup$
    Sorry.I edited it now
    $endgroup$
    – DD90
    Dec 19 '18 at 22:31


















  • $begingroup$
    You mention a $P > 1$ but it is not present anywhere in the series. From the wording, I would expect the sum of $x_n^P$
    $endgroup$
    – Will Jagy
    Dec 19 '18 at 22:30












  • $begingroup$
    Sorry.I edited it now
    $endgroup$
    – DD90
    Dec 19 '18 at 22:31
















$begingroup$
You mention a $P > 1$ but it is not present anywhere in the series. From the wording, I would expect the sum of $x_n^P$
$endgroup$
– Will Jagy
Dec 19 '18 at 22:30






$begingroup$
You mention a $P > 1$ but it is not present anywhere in the series. From the wording, I would expect the sum of $x_n^P$
$endgroup$
– Will Jagy
Dec 19 '18 at 22:30














$begingroup$
Sorry.I edited it now
$endgroup$
– DD90
Dec 19 '18 at 22:31




$begingroup$
Sorry.I edited it now
$endgroup$
– DD90
Dec 19 '18 at 22:31










1 Answer
1






active

oldest

votes


















13












$begingroup$

It suffices to show $x_n < frac{1}{n}$. But it is since $(frac{1}{n})^n+nfrac{1}{n}-1 > 0$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Thank you! so afterwards we can use the convergence of the p- series and the comparison test!
    $endgroup$
    – DD90
    Dec 19 '18 at 22:39











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









13












$begingroup$

It suffices to show $x_n < frac{1}{n}$. But it is since $(frac{1}{n})^n+nfrac{1}{n}-1 > 0$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Thank you! so afterwards we can use the convergence of the p- series and the comparison test!
    $endgroup$
    – DD90
    Dec 19 '18 at 22:39
















13












$begingroup$

It suffices to show $x_n < frac{1}{n}$. But it is since $(frac{1}{n})^n+nfrac{1}{n}-1 > 0$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Thank you! so afterwards we can use the convergence of the p- series and the comparison test!
    $endgroup$
    – DD90
    Dec 19 '18 at 22:39














13












13








13





$begingroup$

It suffices to show $x_n < frac{1}{n}$. But it is since $(frac{1}{n})^n+nfrac{1}{n}-1 > 0$.






share|cite|improve this answer









$endgroup$



It suffices to show $x_n < frac{1}{n}$. But it is since $(frac{1}{n})^n+nfrac{1}{n}-1 > 0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 22:33









mathworker21mathworker21

9,0061928




9,0061928








  • 2




    $begingroup$
    Thank you! so afterwards we can use the convergence of the p- series and the comparison test!
    $endgroup$
    – DD90
    Dec 19 '18 at 22:39














  • 2




    $begingroup$
    Thank you! so afterwards we can use the convergence of the p- series and the comparison test!
    $endgroup$
    – DD90
    Dec 19 '18 at 22:39








2




2




$begingroup$
Thank you! so afterwards we can use the convergence of the p- series and the comparison test!
$endgroup$
– DD90
Dec 19 '18 at 22:39




$begingroup$
Thank you! so afterwards we can use the convergence of the p- series and the comparison test!
$endgroup$
– DD90
Dec 19 '18 at 22:39


















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