Convergence test for a series
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Let S = $sum_{n=1}^{infty}{frac{n^k}{((n^3+n)^{frac{1}{3}}-n)}},forallspacespacespacespace kinmathbb{Z}.$ Say if this series converge of diverge.
My attempt:
$$S=sum_{n=1}^{infty}frac{(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2}{n^{1-k}}$$
Ratio test in undecisive, and Raabe-Duhmal test gave me again $1$ so what test should I use? It is clear that the general term of these series $to0$ iff $1-k<2implies k>1$ and divergent for $k<1$.
sequences-and-series real-numbers numerical-calculus
edited Nov 14 at 11:23
Crazy for maths
4968
asked Nov 14 at 11:09
C. Cristi
1,449218
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up vote
0
down vote
favorite
Let S = $sum_{n=1}^{infty}{frac{n^k}{((n^3+n)^{frac{1}{3}}-n)}},forallspacespacespacespace kinmathbb{Z}.$ Say if this series converge of diverge.
My attempt:
$$S=sum_{n=1}^{infty}frac{(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2}{n^{1-k}}$$
Ratio test in undecisive, and Raabe-Duhmal test gave me again $1$ so what test should I use? It is clear that the general term of these series $to0$ iff $1-k<2implies k>1$ and divergent for $k<1$.
sequences-and-series real-numbers numerical-calculus
edited Nov 14 at 11:23
Crazy for maths
4968
asked Nov 14 at 11:09
C. Cristi
1,449218
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let S = $sum_{n=1}^{infty}{frac{n^k}{((n^3+n)^{frac{1}{3}}-n)}},forallspacespacespacespace kinmathbb{Z}.$ Say if this series converge of diverge.
My attempt:
$$S=sum_{n=1}^{infty}frac{(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2}{n^{1-k}}$$
Ratio test in undecisive, and Raabe-Duhmal test gave me again $1$ so what test should I use? It is clear that the general term of these series $to0$ iff $1-k<2implies k>1$ and divergent for $k<1$.
sequences-and-series real-numbers numerical-calculus
edited Nov 14 at 11:23
Crazy for maths
4968
asked Nov 14 at 11:09
C. Cristi
1,449218
Let S = $sum_{n=1}^{infty}{frac{n^k}{((n^3+n)^{frac{1}{3}}-n)}},forallspacespacespacespace kinmathbb{Z}.$ Say if this series converge of diverge.
My attempt:
$$S=sum_{n=1}^{infty}frac{(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2}{n^{1-k}}$$
Ratio test in undecisive, and Raabe-Duhmal test gave me again $1$ so what test should I use? It is clear that the general term of these series $to0$ iff $1-k<2implies k>1$ and divergent for $k<1$.
sequences-and-series real-numbers numerical-calculus
sequences-and-series real-numbers numerical-calculus
edited Nov 14 at 11:23
Crazy for maths
4968
asked Nov 14 at 11:09
C. Cristi
1,449218
edited Nov 14 at 11:23
Crazy for maths
4968
asked Nov 14 at 11:09
C. Cristi
1,449218
edited Nov 14 at 11:23
Crazy for maths
4968
edited Nov 14 at 11:23
Crazy for maths
4968
edited Nov 14 at 11:23
Crazy for maths
4968
4968
asked Nov 14 at 11:09
C. Cristi
1,449218
asked Nov 14 at 11:09
C. Cristi
1,449218
asked Nov 14 at 11:09
C. Cristi
1,449218
1,449218
add a comment |
add a comment |
2 Answers
2
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up vote
1
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accepted
Note thatbegin{align}frac1{sqrt[3]{n^3+n}-n}&=frac1{sqrt[3]{n^3+n}-sqrt[3]{n^3}}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}{n^3+n-n^3}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}n\&inleft[3n,frac{3sqrt[3]{(n^3+n)^2}}nright].end{align}This is enough to prove that the series converges if and only if $k<-2$.
answered Nov 14 at 11:19
José Carlos Santos
139k18111203
add a comment |
up vote
0
down vote
May be, you could start considering that, for large values of $n$,
$$(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2=3 n^2+1-frac{2}{9 n^2}+Oleft(frac{1}{n^4}right)$$
answered Nov 14 at 11:35
Claude Leibovici
116k1156131
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note thatbegin{align}frac1{sqrt[3]{n^3+n}-n}&=frac1{sqrt[3]{n^3+n}-sqrt[3]{n^3}}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}{n^3+n-n^3}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}n\&inleft[3n,frac{3sqrt[3]{(n^3+n)^2}}nright].end{align}This is enough to prove that the series converges if and only if $k<-2$.
answered Nov 14 at 11:19
José Carlos Santos
139k18111203
add a comment |
up vote
1
down vote
accepted
Note thatbegin{align}frac1{sqrt[3]{n^3+n}-n}&=frac1{sqrt[3]{n^3+n}-sqrt[3]{n^3}}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}{n^3+n-n^3}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}n\&inleft[3n,frac{3sqrt[3]{(n^3+n)^2}}nright].end{align}This is enough to prove that the series converges if and only if $k<-2$.
answered Nov 14 at 11:19
José Carlos Santos
139k18111203
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note thatbegin{align}frac1{sqrt[3]{n^3+n}-n}&=frac1{sqrt[3]{n^3+n}-sqrt[3]{n^3}}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}{n^3+n-n^3}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}n\&inleft[3n,frac{3sqrt[3]{(n^3+n)^2}}nright].end{align}This is enough to prove that the series converges if and only if $k<-2$.
answered Nov 14 at 11:19
José Carlos Santos
139k18111203
Note thatbegin{align}frac1{sqrt[3]{n^3+n}-n}&=frac1{sqrt[3]{n^3+n}-sqrt[3]{n^3}}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}{n^3+n-n^3}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}n\&inleft[3n,frac{3sqrt[3]{(n^3+n)^2}}nright].end{align}This is enough to prove that the series converges if and only if $k<-2$.
answered Nov 14 at 11:19
José Carlos Santos
139k18111203
edited Nov 14 at 11:30
answered Nov 14 at 11:19
José Carlos Santos
139k18111203
answered Nov 14 at 11:19
José Carlos Santos
139k18111203
answered Nov 14 at 11:19
José Carlos Santos
139k18111203
139k18111203
add a comment |
add a comment |
up vote
0
down vote
May be, you could start considering that, for large values of $n$,
$$(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2=3 n^2+1-frac{2}{9 n^2}+Oleft(frac{1}{n^4}right)$$
answered Nov 14 at 11:35
Claude Leibovici
116k1156131
add a comment |
up vote
0
down vote
May be, you could start considering that, for large values of $n$,
$$(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2=3 n^2+1-frac{2}{9 n^2}+Oleft(frac{1}{n^4}right)$$
answered Nov 14 at 11:35
Claude Leibovici
116k1156131
add a comment |
up vote
0
down vote
up vote
0
down vote
May be, you could start considering that, for large values of $n$,
$$(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2=3 n^2+1-frac{2}{9 n^2}+Oleft(frac{1}{n^4}right)$$
answered Nov 14 at 11:35
Claude Leibovici
116k1156131
May be, you could start considering that, for large values of $n$,
$$(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2=3 n^2+1-frac{2}{9 n^2}+Oleft(frac{1}{n^4}right)$$
answered Nov 14 at 11:35
Claude Leibovici
116k1156131
edited Nov 14 at 11:45
answered Nov 14 at 11:35
Claude Leibovici
116k1156131
answered Nov 14 at 11:35
Claude Leibovici
116k1156131
answered Nov 14 at 11:35
Claude Leibovici
116k1156131
116k1156131
add a comment |
add a comment |
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