Convergence of the series $(2x)/(1+x^{2}) + (4x^{3})/(1+x^{4}) + (8x^{7})/(1+x^{8})+ dots , x$ lies in...












0












$begingroup$


The series is -



$(2x)/(1+x^{2}) +(4x^{3})/(1+x^{4}) + (8x^{7})/(1+x^{8})+ dots $
where $x$ lies in $[(-1/2),(1/2)].$



I need to find the interval where it converges.



I couldn't write it in standard form, but i can notice that the numerator is derivative of denominator in each term.



I guess interval of convergence of this series should be similar to the interval of convergence of $$sum_{n=1}^{infty} frac{2n(x)^{(2n-1)}}{1+x^{2n}}$$ which is $(-1,1)$.



I don't know if i am right. Please suggest how to find interval of convergence?






real-analysis







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  • 1




    $begingroup$
    In your equation in the main body, your $left(4xright)$ should be $left(4x^3right)$.
    $endgroup$
    – John Omielan
    Dec 29 '18 at 4:36


















0












$begingroup$


The series is -



$(2x)/(1+x^{2}) +(4x^{3})/(1+x^{4}) + (8x^{7})/(1+x^{8})+ dots $
where $x$ lies in $[(-1/2),(1/2)].$



I need to find the interval where it converges.



I couldn't write it in standard form, but i can notice that the numerator is derivative of denominator in each term.



I guess interval of convergence of this series should be similar to the interval of convergence of $$sum_{n=1}^{infty} frac{2n(x)^{(2n-1)}}{1+x^{2n}}$$ which is $(-1,1)$.



I don't know if i am right. Please suggest how to find interval of convergence?






real-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    In your equation in the main body, your $left(4xright)$ should be $left(4x^3right)$.
    $endgroup$
    – John Omielan
    Dec 29 '18 at 4:36
















0












0








0





$begingroup$


The series is -



$(2x)/(1+x^{2}) +(4x^{3})/(1+x^{4}) + (8x^{7})/(1+x^{8})+ dots $
where $x$ lies in $[(-1/2),(1/2)].$



I need to find the interval where it converges.



I couldn't write it in standard form, but i can notice that the numerator is derivative of denominator in each term.



I guess interval of convergence of this series should be similar to the interval of convergence of $$sum_{n=1}^{infty} frac{2n(x)^{(2n-1)}}{1+x^{2n}}$$ which is $(-1,1)$.



I don't know if i am right. Please suggest how to find interval of convergence?






real-analysis







share|cite|improve this question











$endgroup$




The series is -



$(2x)/(1+x^{2}) +(4x^{3})/(1+x^{4}) + (8x^{7})/(1+x^{8})+ dots $
where $x$ lies in $[(-1/2),(1/2)].$



I need to find the interval where it converges.



I couldn't write it in standard form, but i can notice that the numerator is derivative of denominator in each term.



I guess interval of convergence of this series should be similar to the interval of convergence of $$sum_{n=1}^{infty} frac{2n(x)^{(2n-1)}}{1+x^{2n}}$$ which is $(-1,1)$.



I don't know if i am right. Please suggest how to find interval of convergence?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 4:36







Mathsaddict

















asked Dec 29 '18 at 4:33









MathsaddictMathsaddict

3619




3619








  • 1




    $begingroup$
    In your equation in the main body, your $left(4xright)$ should be $left(4x^3right)$.
    $endgroup$
    – John Omielan
    Dec 29 '18 at 4:36
















  • 1




    $begingroup$
    In your equation in the main body, your $left(4xright)$ should be $left(4x^3right)$.
    $endgroup$
    – John Omielan
    Dec 29 '18 at 4:36










1




1




$begingroup$
In your equation in the main body, your $left(4xright)$ should be $left(4x^3right)$.
$endgroup$
– John Omielan
Dec 29 '18 at 4:36






$begingroup$
In your equation in the main body, your $left(4xright)$ should be $left(4x^3right)$.
$endgroup$
– John Omielan
Dec 29 '18 at 4:36












1 Answer
1






active

oldest

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2












$begingroup$

Use M-test. The series is dominated by $2|x|+4|x|^{3}+8|x|^{7}+cdots$ which is convergent if $x$ lies in $[-frac 1 2, frac 1 2]$. The series $2|x|+4|x|^{3}+8|x|^{7}+cdots$ converges for $|x| <1$ by ratio test, so the given series also converges in $(-1,1)$. It does not converge for $x =pm 1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So the interval of convergence is $(-1,1)$ . How would I check the behaviour of series when $x=1$ or $x=-1$?
    $endgroup$
    – Mathsaddict
    Dec 29 '18 at 6:10












  • $begingroup$
    if x lies in $[-1,1]$
    $endgroup$
    – Mathsaddict
    Dec 29 '18 at 6:31











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Use M-test. The series is dominated by $2|x|+4|x|^{3}+8|x|^{7}+cdots$ which is convergent if $x$ lies in $[-frac 1 2, frac 1 2]$. The series $2|x|+4|x|^{3}+8|x|^{7}+cdots$ converges for $|x| <1$ by ratio test, so the given series also converges in $(-1,1)$. It does not converge for $x =pm 1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So the interval of convergence is $(-1,1)$ . How would I check the behaviour of series when $x=1$ or $x=-1$?
    $endgroup$
    – Mathsaddict
    Dec 29 '18 at 6:10












  • $begingroup$
    if x lies in $[-1,1]$
    $endgroup$
    – Mathsaddict
    Dec 29 '18 at 6:31
















2












$begingroup$

Use M-test. The series is dominated by $2|x|+4|x|^{3}+8|x|^{7}+cdots$ which is convergent if $x$ lies in $[-frac 1 2, frac 1 2]$. The series $2|x|+4|x|^{3}+8|x|^{7}+cdots$ converges for $|x| <1$ by ratio test, so the given series also converges in $(-1,1)$. It does not converge for $x =pm 1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So the interval of convergence is $(-1,1)$ . How would I check the behaviour of series when $x=1$ or $x=-1$?
    $endgroup$
    – Mathsaddict
    Dec 29 '18 at 6:10












  • $begingroup$
    if x lies in $[-1,1]$
    $endgroup$
    – Mathsaddict
    Dec 29 '18 at 6:31














2












2








2





$begingroup$

Use M-test. The series is dominated by $2|x|+4|x|^{3}+8|x|^{7}+cdots$ which is convergent if $x$ lies in $[-frac 1 2, frac 1 2]$. The series $2|x|+4|x|^{3}+8|x|^{7}+cdots$ converges for $|x| <1$ by ratio test, so the given series also converges in $(-1,1)$. It does not converge for $x =pm 1$.






share|cite|improve this answer











$endgroup$



Use M-test. The series is dominated by $2|x|+4|x|^{3}+8|x|^{7}+cdots$ which is convergent if $x$ lies in $[-frac 1 2, frac 1 2]$. The series $2|x|+4|x|^{3}+8|x|^{7}+cdots$ converges for $|x| <1$ by ratio test, so the given series also converges in $(-1,1)$. It does not converge for $x =pm 1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 29 '18 at 11:40

























answered Dec 29 '18 at 4:49









Kavi Rama MurthyKavi Rama Murthy

61.8k42262




61.8k42262












  • $begingroup$
    So the interval of convergence is $(-1,1)$ . How would I check the behaviour of series when $x=1$ or $x=-1$?
    $endgroup$
    – Mathsaddict
    Dec 29 '18 at 6:10












  • $begingroup$
    if x lies in $[-1,1]$
    $endgroup$
    – Mathsaddict
    Dec 29 '18 at 6:31


















  • $begingroup$
    So the interval of convergence is $(-1,1)$ . How would I check the behaviour of series when $x=1$ or $x=-1$?
    $endgroup$
    – Mathsaddict
    Dec 29 '18 at 6:10












  • $begingroup$
    if x lies in $[-1,1]$
    $endgroup$
    – Mathsaddict
    Dec 29 '18 at 6:31
















$begingroup$
So the interval of convergence is $(-1,1)$ . How would I check the behaviour of series when $x=1$ or $x=-1$?
$endgroup$
– Mathsaddict
Dec 29 '18 at 6:10






$begingroup$
So the interval of convergence is $(-1,1)$ . How would I check the behaviour of series when $x=1$ or $x=-1$?
$endgroup$
– Mathsaddict
Dec 29 '18 at 6:10














$begingroup$
if x lies in $[-1,1]$
$endgroup$
– Mathsaddict
Dec 29 '18 at 6:31




$begingroup$
if x lies in $[-1,1]$
$endgroup$
– Mathsaddict
Dec 29 '18 at 6:31


















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