Convergence of the series $(2x)/(1+x^{2}) + (4x^{3})/(1+x^{4}) + (8x^{7})/(1+x^{8})+ dots , x$ lies in...
$begingroup$
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
asked Dec 29 '18 at 4:33
MathsaddictMathsaddict
3619
$endgroup$
add a comment |
$begingroup$
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
asked Dec 29 '18 at 4:33
MathsaddictMathsaddict
3619
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 29 '18 at 4:33
MathsaddictMathsaddict
3619
$endgroup$
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
asked Dec 29 '18 at 4:33
MathsaddictMathsaddict
3619
asked Dec 29 '18 at 4:33
MathsaddictMathsaddict
3619
edited Dec 29 '18 at 4:36
Mathsaddict
asked Dec 29 '18 at 4:33
MathsaddictMathsaddict
3619
asked Dec 29 '18 at 4:33
MathsaddictMathsaddict
3619
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $x =pm 1$.
answered Dec 29 '18 at 4:49
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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$begingroup$
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 $x =pm 1$.
answered Dec 29 '18 at 4:49
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $x =pm 1$.
answered Dec 29 '18 at 4:49
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $x =pm 1$.
answered Dec 29 '18 at 4:49
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
$endgroup$
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 $x =pm 1$.
answered Dec 29 '18 at 4:49
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
edited Dec 29 '18 at 11:40
answered Dec 29 '18 at 4:49
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
answered Dec 29 '18 at 4:49
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
In your equation in the main body, your $left(4xright)$ should be $left(4x^3right)$.
$endgroup$
– John Omielan
Dec 29 '18 at 4:36