Converge or diverge? $sum_{n=1}^{infty}frac {(ln n)^a}{n^b}$
up vote
0
down vote
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Converge or diverge? $sum_{n=1}^{infty}frac {(ln n)^a}{n^b}$, $a > 0, b > 1$.
How do you approach this? Since every criteria I know gives $1$. What should I do?
$lim_{ntoinfty} frac {a_{n+1}}{a_n} = 1.$
$lim_{ntoinfty} (a_n)^{frac 1n} = 1.$
I think this should converge for every $a,b$ with the restriction in hypothesis, right?
sequences-and-series convergence
edited Nov 27 at 16:36
amWhy
191k28223439
asked Nov 15 at 15:18
C. Cristi
1,471218
add a comment |
up vote
0
down vote
favorite
Converge or diverge? $sum_{n=1}^{infty}frac {(ln n)^a}{n^b}$, $a > 0, b > 1$.
How do you approach this? Since every criteria I know gives $1$. What should I do?
$lim_{ntoinfty} frac {a_{n+1}}{a_n} = 1.$
$lim_{ntoinfty} (a_n)^{frac 1n} = 1.$
I think this should converge for every $a,b$ with the restriction in hypothesis, right?
sequences-and-series convergence
edited Nov 27 at 16:36
amWhy
191k28223439
asked Nov 15 at 15:18
C. Cristi
1,471218
Hint: For large enough $n$ we have $(ln n)^a<n^{(b-1)/2}$.
– Jyrki Lahtonen
Nov 19 at 17:11
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Converge or diverge? $sum_{n=1}^{infty}frac {(ln n)^a}{n^b}$, $a > 0, b > 1$.
How do you approach this? Since every criteria I know gives $1$. What should I do?
$lim_{ntoinfty} frac {a_{n+1}}{a_n} = 1.$
$lim_{ntoinfty} (a_n)^{frac 1n} = 1.$
I think this should converge for every $a,b$ with the restriction in hypothesis, right?
sequences-and-series convergence
edited Nov 27 at 16:36
amWhy
191k28223439
asked Nov 15 at 15:18
C. Cristi
1,471218
Converge or diverge? $sum_{n=1}^{infty}frac {(ln n)^a}{n^b}$, $a > 0, b > 1$.
How do you approach this? Since every criteria I know gives $1$. What should I do?
$lim_{ntoinfty} frac {a_{n+1}}{a_n} = 1.$
$lim_{ntoinfty} (a_n)^{frac 1n} = 1.$
I think this should converge for every $a,b$ with the restriction in hypothesis, right?
sequences-and-series convergence
sequences-and-series convergence
edited Nov 27 at 16:36
amWhy
191k28223439
asked Nov 15 at 15:18
C. Cristi
1,471218
edited Nov 27 at 16:36
amWhy
191k28223439
asked Nov 15 at 15:18
C. Cristi
1,471218
edited Nov 27 at 16:36
amWhy
191k28223439
edited Nov 27 at 16:36
amWhy
191k28223439
edited Nov 27 at 16:36
amWhy
191k28223439
191k28223439
asked Nov 15 at 15:18
C. Cristi
1,471218
asked Nov 15 at 15:18
C. Cristi
1,471218
asked Nov 15 at 15:18
C. Cristi
1,471218
1,471218
Hint: For large enough $n$ we have $(ln n)^a<n^{(b-1)/2}$.
– Jyrki Lahtonen
Nov 19 at 17:11
add a comment |
Hint: For large enough $n$ we have $(ln n)^a<n^{(b-1)/2}$.
– Jyrki Lahtonen
Nov 19 at 17:11
Hint: For large enough $n$ we have $(ln n)^a<n^{(b-1)/2}$.
– Jyrki Lahtonen
Nov 19 at 17:11
Hint: For large enough $n$ we have $(ln n)^a<n^{(b-1)/2}$.
– Jyrki Lahtonen
Nov 19 at 17:11
add a comment |
3 Answers
3
active
oldest
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up vote
1
down vote
Hint Apply the integral test, that is, consider the convergence of $$int_1^{infty} frac{log^a x ,dx}{x^b} = int_0^{infty} u^a e^{(1 - b) u} du .$$
answered Nov 15 at 15:29
Travis
59.1k766144
is that looking the same as Gamma Function?
– C. Cristi
Nov 19 at 15:34
It can be transformed into an expression involving the Gamma function, yes.
– Travis
Nov 19 at 22:28
(...but one doesn't need to know that to check convergence.)
– Travis
Nov 19 at 22:29
add a comment |
up vote
1
down vote
Write $b=1+2ac$ with $cgt0$. Then show that
$${(ln n)^a/n^{1+2ac}over1/n^{1+ac}}=left(ln nover n^cright)^ato0$$
Remarks (added later): What we're really after is an inequality of the form
$${(ln n)^aover n^b}le{Bover n^{1+p}}$$
for some $pgt0$, in which case the convergence of $sum{1over n^{1+p}}$ guarantees the convergence of $sum{(ln n)^aover n^b}$. So to be explicit, if we write $b=1+2ac$ and $p=ac$ (with $c={b-1over2a}gt0$), then we have
$${(ln n)^a/n^bover1/n^{1+p}}={(ln n)^a/n^{1+2ac}over1/n^{1+ac}}=left(ln nover n^cright)^a=left(ln n^cover cn^cright)^aleleft(1over ce right)^a$$
where we use the fact that $f(x)={ln xover x}$ attains its maximum at $x=e$, where $f(e)={1over e}$. It follows that
$$sum_{n=1}^infty{(ln n)^aover n^b}leleft(1over ce right)^asum_{n=1}^infty{1over n^{1+ac}}ltinfty$$
where $c={b-1over2a}$. Writing it without the $c$, it's
$$sum_{n=1}^infty{(ln n)^aover n^b}leleft(2aover (b-1)e right)^asum_{n=1}^infty{1over n^{1+(b-1)/2}}ltinfty$$
answered Nov 15 at 15:25
Barry Cipra
58.6k653122
Woah, woah, slow down, why did you wrote it like that? to get the same power? and also why you need the $c$? Why the $2$? and what is the intuition behind that?
– C. Cristi
Nov 15 at 15:27
And why it is enough to show that it tends to $0$?
– C. Cristi
Nov 15 at 15:28
I wrote it that way just to make things look nice. It's actually enough to show that the ratio is ultimately bounded, at which point ${(ln n)^aover n^{1+2ac}}lt{Bover n^{1+ac}}$ shows that given series is bounded by a convergent series.
– Barry Cipra
Nov 15 at 17:02
I don't think that's enough for me, can you be more explicit?
– C. Cristi
Nov 19 at 15:34
add a comment |
up vote
0
down vote
By Cauchy condensation test the condensed series is
$$sum_{n=1}^{infty}frac {2^n(ln 2^n)^a}{2^{bn}}=sum_{n=1}^{infty}frac {n^aln^a 2}{2^{n(b-1)}}$$
which converges for $b>1$.
answered Nov 19 at 16:48
gimusi
91.7k84495
add a comment |
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3 Answers
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3 Answers
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active
oldest
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active
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up vote
1
down vote
Hint Apply the integral test, that is, consider the convergence of $$int_1^{infty} frac{log^a x ,dx}{x^b} = int_0^{infty} u^a e^{(1 - b) u} du .$$
answered Nov 15 at 15:29
Travis
59.1k766144
is that looking the same as Gamma Function?
– C. Cristi
Nov 19 at 15:34
It can be transformed into an expression involving the Gamma function, yes.
– Travis
Nov 19 at 22:28
(...but one doesn't need to know that to check convergence.)
– Travis
Nov 19 at 22:29
add a comment |
up vote
1
down vote
Hint Apply the integral test, that is, consider the convergence of $$int_1^{infty} frac{log^a x ,dx}{x^b} = int_0^{infty} u^a e^{(1 - b) u} du .$$
answered Nov 15 at 15:29
Travis
59.1k766144
is that looking the same as Gamma Function?
– C. Cristi
Nov 19 at 15:34
It can be transformed into an expression involving the Gamma function, yes.
– Travis
Nov 19 at 22:28
(...but one doesn't need to know that to check convergence.)
– Travis
Nov 19 at 22:29
add a comment |
up vote
1
down vote
up vote
1
down vote
Hint Apply the integral test, that is, consider the convergence of $$int_1^{infty} frac{log^a x ,dx}{x^b} = int_0^{infty} u^a e^{(1 - b) u} du .$$
answered Nov 15 at 15:29
Travis
59.1k766144
Hint Apply the integral test, that is, consider the convergence of $$int_1^{infty} frac{log^a x ,dx}{x^b} = int_0^{infty} u^a e^{(1 - b) u} du .$$
answered Nov 15 at 15:29
Travis
59.1k766144
answered Nov 15 at 15:29
Travis
59.1k766144
answered Nov 15 at 15:29
Travis
59.1k766144
answered Nov 15 at 15:29
Travis
59.1k766144
59.1k766144
is that looking the same as Gamma Function?
– C. Cristi
Nov 19 at 15:34
It can be transformed into an expression involving the Gamma function, yes.
– Travis
Nov 19 at 22:28
(...but one doesn't need to know that to check convergence.)
– Travis
Nov 19 at 22:29
add a comment |
is that looking the same as Gamma Function?
– C. Cristi
Nov 19 at 15:34
It can be transformed into an expression involving the Gamma function, yes.
– Travis
Nov 19 at 22:28
(...but one doesn't need to know that to check convergence.)
– Travis
Nov 19 at 22:29
is that looking the same as Gamma Function?
– C. Cristi
Nov 19 at 15:34
is that looking the same as Gamma Function?
– C. Cristi
Nov 19 at 15:34
It can be transformed into an expression involving the Gamma function, yes.
– Travis
Nov 19 at 22:28
It can be transformed into an expression involving the Gamma function, yes.
– Travis
Nov 19 at 22:28
(...but one doesn't need to know that to check convergence.)
– Travis
Nov 19 at 22:29
(...but one doesn't need to know that to check convergence.)
– Travis
Nov 19 at 22:29
add a comment |
up vote
1
down vote
Write $b=1+2ac$ with $cgt0$. Then show that
$${(ln n)^a/n^{1+2ac}over1/n^{1+ac}}=left(ln nover n^cright)^ato0$$
Remarks (added later): What we're really after is an inequality of the form
$${(ln n)^aover n^b}le{Bover n^{1+p}}$$
for some $pgt0$, in which case the convergence of $sum{1over n^{1+p}}$ guarantees the convergence of $sum{(ln n)^aover n^b}$. So to be explicit, if we write $b=1+2ac$ and $p=ac$ (with $c={b-1over2a}gt0$), then we have
$${(ln n)^a/n^bover1/n^{1+p}}={(ln n)^a/n^{1+2ac}over1/n^{1+ac}}=left(ln nover n^cright)^a=left(ln n^cover cn^cright)^aleleft(1over ce right)^a$$
where we use the fact that $f(x)={ln xover x}$ attains its maximum at $x=e$, where $f(e)={1over e}$. It follows that
$$sum_{n=1}^infty{(ln n)^aover n^b}leleft(1over ce right)^asum_{n=1}^infty{1over n^{1+ac}}ltinfty$$
where $c={b-1over2a}$. Writing it without the $c$, it's
$$sum_{n=1}^infty{(ln n)^aover n^b}leleft(2aover (b-1)e right)^asum_{n=1}^infty{1over n^{1+(b-1)/2}}ltinfty$$
answered Nov 15 at 15:25
Barry Cipra
58.6k653122
Woah, woah, slow down, why did you wrote it like that? to get the same power? and also why you need the $c$? Why the $2$? and what is the intuition behind that?
– C. Cristi
Nov 15 at 15:27
And why it is enough to show that it tends to $0$?
– C. Cristi
Nov 15 at 15:28
I wrote it that way just to make things look nice. It's actually enough to show that the ratio is ultimately bounded, at which point ${(ln n)^aover n^{1+2ac}}lt{Bover n^{1+ac}}$ shows that given series is bounded by a convergent series.
– Barry Cipra
Nov 15 at 17:02
I don't think that's enough for me, can you be more explicit?
– C. Cristi
Nov 19 at 15:34
add a comment |
up vote
1
down vote
Write $b=1+2ac$ with $cgt0$. Then show that
$${(ln n)^a/n^{1+2ac}over1/n^{1+ac}}=left(ln nover n^cright)^ato0$$
Remarks (added later): What we're really after is an inequality of the form
$${(ln n)^aover n^b}le{Bover n^{1+p}}$$
for some $pgt0$, in which case the convergence of $sum{1over n^{1+p}}$ guarantees the convergence of $sum{(ln n)^aover n^b}$. So to be explicit, if we write $b=1+2ac$ and $p=ac$ (with $c={b-1over2a}gt0$), then we have
$${(ln n)^a/n^bover1/n^{1+p}}={(ln n)^a/n^{1+2ac}over1/n^{1+ac}}=left(ln nover n^cright)^a=left(ln n^cover cn^cright)^aleleft(1over ce right)^a$$
where we use the fact that $f(x)={ln xover x}$ attains its maximum at $x=e$, where $f(e)={1over e}$. It follows that
$$sum_{n=1}^infty{(ln n)^aover n^b}leleft(1over ce right)^asum_{n=1}^infty{1over n^{1+ac}}ltinfty$$
where $c={b-1over2a}$. Writing it without the $c$, it's
$$sum_{n=1}^infty{(ln n)^aover n^b}leleft(2aover (b-1)e right)^asum_{n=1}^infty{1over n^{1+(b-1)/2}}ltinfty$$
answered Nov 15 at 15:25
Barry Cipra
58.6k653122
Woah, woah, slow down, why did you wrote it like that? to get the same power? and also why you need the $c$? Why the $2$? and what is the intuition behind that?
– C. Cristi
Nov 15 at 15:27
And why it is enough to show that it tends to $0$?
– C. Cristi
Nov 15 at 15:28
I wrote it that way just to make things look nice. It's actually enough to show that the ratio is ultimately bounded, at which point ${(ln n)^aover n^{1+2ac}}lt{Bover n^{1+ac}}$ shows that given series is bounded by a convergent series.
– Barry Cipra
Nov 15 at 17:02
I don't think that's enough for me, can you be more explicit?
– C. Cristi
Nov 19 at 15:34
add a comment |
up vote
1
down vote
up vote
1
down vote
Write $b=1+2ac$ with $cgt0$. Then show that
$${(ln n)^a/n^{1+2ac}over1/n^{1+ac}}=left(ln nover n^cright)^ato0$$
Remarks (added later): What we're really after is an inequality of the form
$${(ln n)^aover n^b}le{Bover n^{1+p}}$$
for some $pgt0$, in which case the convergence of $sum{1over n^{1+p}}$ guarantees the convergence of $sum{(ln n)^aover n^b}$. So to be explicit, if we write $b=1+2ac$ and $p=ac$ (with $c={b-1over2a}gt0$), then we have
$${(ln n)^a/n^bover1/n^{1+p}}={(ln n)^a/n^{1+2ac}over1/n^{1+ac}}=left(ln nover n^cright)^a=left(ln n^cover cn^cright)^aleleft(1over ce right)^a$$
where we use the fact that $f(x)={ln xover x}$ attains its maximum at $x=e$, where $f(e)={1over e}$. It follows that
$$sum_{n=1}^infty{(ln n)^aover n^b}leleft(1over ce right)^asum_{n=1}^infty{1over n^{1+ac}}ltinfty$$
where $c={b-1over2a}$. Writing it without the $c$, it's
$$sum_{n=1}^infty{(ln n)^aover n^b}leleft(2aover (b-1)e right)^asum_{n=1}^infty{1over n^{1+(b-1)/2}}ltinfty$$
answered Nov 15 at 15:25
Barry Cipra
58.6k653122
Write $b=1+2ac$ with $cgt0$. Then show that
$${(ln n)^a/n^{1+2ac}over1/n^{1+ac}}=left(ln nover n^cright)^ato0$$
Remarks (added later): What we're really after is an inequality of the form
$${(ln n)^aover n^b}le{Bover n^{1+p}}$$
for some $pgt0$, in which case the convergence of $sum{1over n^{1+p}}$ guarantees the convergence of $sum{(ln n)^aover n^b}$. So to be explicit, if we write $b=1+2ac$ and $p=ac$ (with $c={b-1over2a}gt0$), then we have
$${(ln n)^a/n^bover1/n^{1+p}}={(ln n)^a/n^{1+2ac}over1/n^{1+ac}}=left(ln nover n^cright)^a=left(ln n^cover cn^cright)^aleleft(1over ce right)^a$$
where we use the fact that $f(x)={ln xover x}$ attains its maximum at $x=e$, where $f(e)={1over e}$. It follows that
$$sum_{n=1}^infty{(ln n)^aover n^b}leleft(1over ce right)^asum_{n=1}^infty{1over n^{1+ac}}ltinfty$$
where $c={b-1over2a}$. Writing it without the $c$, it's
$$sum_{n=1}^infty{(ln n)^aover n^b}leleft(2aover (b-1)e right)^asum_{n=1}^infty{1over n^{1+(b-1)/2}}ltinfty$$
answered Nov 15 at 15:25
Barry Cipra
58.6k653122
edited Nov 19 at 16:39
answered Nov 15 at 15:25
Barry Cipra
58.6k653122
answered Nov 15 at 15:25
Barry Cipra
58.6k653122
answered Nov 15 at 15:25
Barry Cipra
58.6k653122
58.6k653122
Woah, woah, slow down, why did you wrote it like that? to get the same power? and also why you need the $c$? Why the $2$? and what is the intuition behind that?
– C. Cristi
Nov 15 at 15:27
And why it is enough to show that it tends to $0$?
– C. Cristi
Nov 15 at 15:28
I wrote it that way just to make things look nice. It's actually enough to show that the ratio is ultimately bounded, at which point ${(ln n)^aover n^{1+2ac}}lt{Bover n^{1+ac}}$ shows that given series is bounded by a convergent series.
– Barry Cipra
Nov 15 at 17:02
I don't think that's enough for me, can you be more explicit?
– C. Cristi
Nov 19 at 15:34
add a comment |
Woah, woah, slow down, why did you wrote it like that? to get the same power? and also why you need the $c$? Why the $2$? and what is the intuition behind that?
– C. Cristi
Nov 15 at 15:27
And why it is enough to show that it tends to $0$?
– C. Cristi
Nov 15 at 15:28
I wrote it that way just to make things look nice. It's actually enough to show that the ratio is ultimately bounded, at which point ${(ln n)^aover n^{1+2ac}}lt{Bover n^{1+ac}}$ shows that given series is bounded by a convergent series.
– Barry Cipra
Nov 15 at 17:02
I don't think that's enough for me, can you be more explicit?
– C. Cristi
Nov 19 at 15:34
Woah, woah, slow down, why did you wrote it like that? to get the same power? and also why you need the $c$? Why the $2$? and what is the intuition behind that?
– C. Cristi
Nov 15 at 15:27
Woah, woah, slow down, why did you wrote it like that? to get the same power? and also why you need the $c$? Why the $2$? and what is the intuition behind that?
– C. Cristi
Nov 15 at 15:27
And why it is enough to show that it tends to $0$?
– C. Cristi
Nov 15 at 15:28
And why it is enough to show that it tends to $0$?
– C. Cristi
Nov 15 at 15:28
I wrote it that way just to make things look nice. It's actually enough to show that the ratio is ultimately bounded, at which point ${(ln n)^aover n^{1+2ac}}lt{Bover n^{1+ac}}$ shows that given series is bounded by a convergent series.
– Barry Cipra
Nov 15 at 17:02
I wrote it that way just to make things look nice. It's actually enough to show that the ratio is ultimately bounded, at which point ${(ln n)^aover n^{1+2ac}}lt{Bover n^{1+ac}}$ shows that given series is bounded by a convergent series.
– Barry Cipra
Nov 15 at 17:02
I don't think that's enough for me, can you be more explicit?
– C. Cristi
Nov 19 at 15:34
I don't think that's enough for me, can you be more explicit?
– C. Cristi
Nov 19 at 15:34
add a comment |
up vote
0
down vote
By Cauchy condensation test the condensed series is
$$sum_{n=1}^{infty}frac {2^n(ln 2^n)^a}{2^{bn}}=sum_{n=1}^{infty}frac {n^aln^a 2}{2^{n(b-1)}}$$
which converges for $b>1$.
answered Nov 19 at 16:48
gimusi
91.7k84495
add a comment |
up vote
0
down vote
By Cauchy condensation test the condensed series is
$$sum_{n=1}^{infty}frac {2^n(ln 2^n)^a}{2^{bn}}=sum_{n=1}^{infty}frac {n^aln^a 2}{2^{n(b-1)}}$$
which converges for $b>1$.
answered Nov 19 at 16:48
gimusi
91.7k84495
add a comment |
up vote
0
down vote
up vote
0
down vote
By Cauchy condensation test the condensed series is
$$sum_{n=1}^{infty}frac {2^n(ln 2^n)^a}{2^{bn}}=sum_{n=1}^{infty}frac {n^aln^a 2}{2^{n(b-1)}}$$
which converges for $b>1$.
answered Nov 19 at 16:48
gimusi
91.7k84495
By Cauchy condensation test the condensed series is
$$sum_{n=1}^{infty}frac {2^n(ln 2^n)^a}{2^{bn}}=sum_{n=1}^{infty}frac {n^aln^a 2}{2^{n(b-1)}}$$
which converges for $b>1$.
answered Nov 19 at 16:48
gimusi
91.7k84495
edited Nov 19 at 16:56
answered Nov 19 at 16:48
gimusi
91.7k84495
answered Nov 19 at 16:48
gimusi
91.7k84495
answered Nov 19 at 16:48
gimusi
91.7k84495
91.7k84495
add a comment |
add a comment |
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Hint: For large enough $n$ we have $(ln n)^a<n^{(b-1)/2}$.
– Jyrki Lahtonen
Nov 19 at 17:11