$sum_{n=0}^infty (sqrt[3]{n+1} - sqrt[3]{n})^alpha (2+n^4)^beta$ convergence/divergence
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I am trying to analyze the convergence/divergence of:
$sum_{n=0}^infty (sqrt[3]{n+1} - sqrt[3]{n})^alpha (2+n^4)^beta$
depending on $alpha$ and $beta$. This is how far I have gone:
First i have changed it a bit using $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ and i have studied this: $sum_{n=0}^infty frac{(2+n^4)^beta}{((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}$ . I used the limit-convergence test
$lim_{nto infty} frac{(2+n^4)^beta}{n^lambda((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}=l$ and now I would have to chose $lambda$ so that $0lt l lt infty$. My problem comes now. If i choose $lambda= 4beta-frac{2}{3}alpha$ the limit shoud be finite and bigger than $0$. So now, would it be correct to say that if $-1 gt 4beta-frac{2}{3}alpha$ the series is convergent? If it is not correct, how should the problem be done? Thanks in advance.
sequences-and-series
asked Dec 24 '18 at 20:20
AndarrkorAndarrkor
466
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add a comment |
$begingroup$
I am trying to analyze the convergence/divergence of:
$sum_{n=0}^infty (sqrt[3]{n+1} - sqrt[3]{n})^alpha (2+n^4)^beta$
depending on $alpha$ and $beta$. This is how far I have gone:
First i have changed it a bit using $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ and i have studied this: $sum_{n=0}^infty frac{(2+n^4)^beta}{((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}$ . I used the limit-convergence test
$lim_{nto infty} frac{(2+n^4)^beta}{n^lambda((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}=l$ and now I would have to chose $lambda$ so that $0lt l lt infty$. My problem comes now. If i choose $lambda= 4beta-frac{2}{3}alpha$ the limit shoud be finite and bigger than $0$. So now, would it be correct to say that if $-1 gt 4beta-frac{2}{3}alpha$ the series is convergent? If it is not correct, how should the problem be done? Thanks in advance.
sequences-and-series
asked Dec 24 '18 at 20:20
AndarrkorAndarrkor
466
$endgroup$
$begingroup$
@Clayton $a=sqrt[3]{n+1}$ and $b=sqrt[3]{n}$
$endgroup$
– Andarrkor
Dec 24 '18 at 20:57
add a comment |
$begingroup$
I am trying to analyze the convergence/divergence of:
$sum_{n=0}^infty (sqrt[3]{n+1} - sqrt[3]{n})^alpha (2+n^4)^beta$
depending on $alpha$ and $beta$. This is how far I have gone:
First i have changed it a bit using $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ and i have studied this: $sum_{n=0}^infty frac{(2+n^4)^beta}{((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}$ . I used the limit-convergence test
$lim_{nto infty} frac{(2+n^4)^beta}{n^lambda((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}=l$ and now I would have to chose $lambda$ so that $0lt l lt infty$. My problem comes now. If i choose $lambda= 4beta-frac{2}{3}alpha$ the limit shoud be finite and bigger than $0$. So now, would it be correct to say that if $-1 gt 4beta-frac{2}{3}alpha$ the series is convergent? If it is not correct, how should the problem be done? Thanks in advance.
sequences-and-series
asked Dec 24 '18 at 20:20
AndarrkorAndarrkor
466
$endgroup$
I am trying to analyze the convergence/divergence of:
$sum_{n=0}^infty (sqrt[3]{n+1} - sqrt[3]{n})^alpha (2+n^4)^beta$
depending on $alpha$ and $beta$. This is how far I have gone:
First i have changed it a bit using $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ and i have studied this: $sum_{n=0}^infty frac{(2+n^4)^beta}{((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}$ . I used the limit-convergence test
$lim_{nto infty} frac{(2+n^4)^beta}{n^lambda((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}=l$ and now I would have to chose $lambda$ so that $0lt l lt infty$. My problem comes now. If i choose $lambda= 4beta-frac{2}{3}alpha$ the limit shoud be finite and bigger than $0$. So now, would it be correct to say that if $-1 gt 4beta-frac{2}{3}alpha$ the series is convergent? If it is not correct, how should the problem be done? Thanks in advance.
sequences-and-series
sequences-and-series
asked Dec 24 '18 at 20:20
AndarrkorAndarrkor
466
asked Dec 24 '18 at 20:20
AndarrkorAndarrkor
466
asked Dec 24 '18 at 20:20
AndarrkorAndarrkor
466
asked Dec 24 '18 at 20:20
AndarrkorAndarrkor
466
asked Dec 24 '18 at 20:20
AndarrkorAndarrkor
466
466
$begingroup$
@Clayton $a=sqrt[3]{n+1}$ and $b=sqrt[3]{n}$
$endgroup$
– Andarrkor
Dec 24 '18 at 20:57
add a comment |
$begingroup$
@Clayton $a=sqrt[3]{n+1}$ and $b=sqrt[3]{n}$
$endgroup$
– Andarrkor
Dec 24 '18 at 20:57
$begingroup$
@Clayton $a=sqrt[3]{n+1}$ and $b=sqrt[3]{n}$
$endgroup$
– Andarrkor
Dec 24 '18 at 20:57
$begingroup$
@Clayton $a=sqrt[3]{n+1}$ and $b=sqrt[3]{n}$
$endgroup$
– Andarrkor
Dec 24 '18 at 20:57
add a comment |
1 Answer
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It's much simpler using equivalents
$sqrt[3]{n+1}-sqrt[3]{n}=sqrt[3]{n}Bigl(sqrt[3]{1+dfrac 1n}-1Bigr)=sqrt[3]{n}dfrac1{3n}=dfrac1{3n^{tfrac23}}$, so
$$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alphasim_{infty}dfrac1{3^alpha,n^{tfrac{2alpha}3}}.$$
$(2+n^4)^beta sim_infty n^{4beta}.$
Therefore,
$$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alpha (2+n^4)^betasim_infty dfrac1{3^alpha,n^{tfrac{2alpha}3-4beta}}$$
and the latter, which is a $p$-series, converges if and only if $;dfrac{2alpha}3-4beta>1$.
answered Dec 24 '18 at 21:26
BernardBernard
125k743119
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1 Answer
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1 Answer
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active
oldest
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active
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$begingroup$
It's much simpler using equivalents
$sqrt[3]{n+1}-sqrt[3]{n}=sqrt[3]{n}Bigl(sqrt[3]{1+dfrac 1n}-1Bigr)=sqrt[3]{n}dfrac1{3n}=dfrac1{3n^{tfrac23}}$, so
$$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alphasim_{infty}dfrac1{3^alpha,n^{tfrac{2alpha}3}}.$$
$(2+n^4)^beta sim_infty n^{4beta}.$
Therefore,
$$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alpha (2+n^4)^betasim_infty dfrac1{3^alpha,n^{tfrac{2alpha}3-4beta}}$$
and the latter, which is a $p$-series, converges if and only if $;dfrac{2alpha}3-4beta>1$.
answered Dec 24 '18 at 21:26
BernardBernard
125k743119
$endgroup$
add a comment |
$begingroup$
It's much simpler using equivalents
$sqrt[3]{n+1}-sqrt[3]{n}=sqrt[3]{n}Bigl(sqrt[3]{1+dfrac 1n}-1Bigr)=sqrt[3]{n}dfrac1{3n}=dfrac1{3n^{tfrac23}}$, so
$$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alphasim_{infty}dfrac1{3^alpha,n^{tfrac{2alpha}3}}.$$
$(2+n^4)^beta sim_infty n^{4beta}.$
Therefore,
$$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alpha (2+n^4)^betasim_infty dfrac1{3^alpha,n^{tfrac{2alpha}3-4beta}}$$
and the latter, which is a $p$-series, converges if and only if $;dfrac{2alpha}3-4beta>1$.
answered Dec 24 '18 at 21:26
BernardBernard
125k743119
$endgroup$
add a comment |
$begingroup$
It's much simpler using equivalents
$sqrt[3]{n+1}-sqrt[3]{n}=sqrt[3]{n}Bigl(sqrt[3]{1+dfrac 1n}-1Bigr)=sqrt[3]{n}dfrac1{3n}=dfrac1{3n^{tfrac23}}$, so
$$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alphasim_{infty}dfrac1{3^alpha,n^{tfrac{2alpha}3}}.$$
$(2+n^4)^beta sim_infty n^{4beta}.$
Therefore,
$$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alpha (2+n^4)^betasim_infty dfrac1{3^alpha,n^{tfrac{2alpha}3-4beta}}$$
and the latter, which is a $p$-series, converges if and only if $;dfrac{2alpha}3-4beta>1$.
answered Dec 24 '18 at 21:26
BernardBernard
125k743119
$endgroup$
It's much simpler using equivalents
$sqrt[3]{n+1}-sqrt[3]{n}=sqrt[3]{n}Bigl(sqrt[3]{1+dfrac 1n}-1Bigr)=sqrt[3]{n}dfrac1{3n}=dfrac1{3n^{tfrac23}}$, so
$$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alphasim_{infty}dfrac1{3^alpha,n^{tfrac{2alpha}3}}.$$
$(2+n^4)^beta sim_infty n^{4beta}.$
Therefore,
$$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alpha (2+n^4)^betasim_infty dfrac1{3^alpha,n^{tfrac{2alpha}3-4beta}}$$
and the latter, which is a $p$-series, converges if and only if $;dfrac{2alpha}3-4beta>1$.
answered Dec 24 '18 at 21:26
BernardBernard
125k743119
answered Dec 24 '18 at 21:26
BernardBernard
125k743119
answered Dec 24 '18 at 21:26
BernardBernard
125k743119
answered Dec 24 '18 at 21:26
BernardBernard
125k743119
125k743119
add a comment |
add a comment |
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$begingroup$
@Clayton $a=sqrt[3]{n+1}$ and $b=sqrt[3]{n}$
$endgroup$
– Andarrkor
Dec 24 '18 at 20:57