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Depending on the parameter $alpha$ study the convergence of the series $sum_n=2^inftyfraclog(logn)log^alphan$
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Depending on the parameter $alpha$ study the convergence of the series $$sum_n=2^inftyfraclog(logn)log^alphan$$
So for $alpha gt 0$ and $x gt 1$ the function $f(x)=fraclog(logx)log^alphax$ is decreasing so I can use the integral test
which is divergent for $alpha =1$
Is my reasoning right? I don't know what to do next
calculus sequences-and-series convergence
asked Jul 27 at 6:41
J.Dane
159112
add a comment |Â
up vote
1
down vote
favorite
Depending on the parameter $alpha$ study the convergence of the series $$sum_n=2^inftyfraclog(logn)log^alphan$$
So for $alpha gt 0$ and $x gt 1$ the function $f(x)=fraclog(logx)log^alphax$ is decreasing so I can use the integral test
which is divergent for $alpha =1$
Is my reasoning right? I don't know what to do next
calculus sequences-and-series convergence
asked Jul 27 at 6:41
J.Dane
159112
How do you prove that the integral diverges?
– gimusi
Jul 27 at 6:48
I fixed it. I hope it's right
– J.Dane
Jul 27 at 7:03
It seems uncorrect, try to derive to check. Cauchy condensation test would simplify a lot.
– gimusi
Jul 27 at 7:09
What is $int u,mathrmdu$? It's not $frac u2+C$. Even when that's fixed, what about $alphane1$?
– robjohn♦
Jul 27 at 7:54
Yes, I had my integral wrong.
– J.Dane
Jul 27 at 7:56
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Depending on the parameter $alpha$ study the convergence of the series $$sum_n=2^inftyfraclog(logn)log^alphan$$
So for $alpha gt 0$ and $x gt 1$ the function $f(x)=fraclog(logx)log^alphax$ is decreasing so I can use the integral test
which is divergent for $alpha =1$
Is my reasoning right? I don't know what to do next
calculus sequences-and-series convergence
asked Jul 27 at 6:41
J.Dane
159112
Depending on the parameter $alpha$ study the convergence of the series $$sum_n=2^inftyfraclog(logn)log^alphan$$
So for $alpha gt 0$ and $x gt 1$ the function $f(x)=fraclog(logx)log^alphax$ is decreasing so I can use the integral test
which is divergent for $alpha =1$
Is my reasoning right? I don't know what to do next
calculus sequences-and-series convergence
asked Jul 27 at 6:41
J.Dane
159112
edited Jul 27 at 7:55
asked Jul 27 at 6:41
J.Dane
159112
asked Jul 27 at 6:41
J.Dane
159112
asked Jul 27 at 6:41
J.Dane
159112
159112
How do you prove that the integral diverges?
– gimusi
Jul 27 at 6:48
I fixed it. I hope it's right
– J.Dane
Jul 27 at 7:03
It seems uncorrect, try to derive to check. Cauchy condensation test would simplify a lot.
– gimusi
Jul 27 at 7:09
What is $int u,mathrmdu$? It's not $frac u2+C$. Even when that's fixed, what about $alphane1$?
– robjohn♦
Jul 27 at 7:54
Yes, I had my integral wrong.
– J.Dane
Jul 27 at 7:56
add a comment |Â
How do you prove that the integral diverges?
– gimusi
Jul 27 at 6:48
I fixed it. I hope it's right
– J.Dane
Jul 27 at 7:03
It seems uncorrect, try to derive to check. Cauchy condensation test would simplify a lot.
– gimusi
Jul 27 at 7:09
What is $int u,mathrmdu$? It's not $frac u2+C$. Even when that's fixed, what about $alphane1$?
– robjohn♦
Jul 27 at 7:54
Yes, I had my integral wrong.
– J.Dane
Jul 27 at 7:56
How do you prove that the integral diverges?
– gimusi
Jul 27 at 6:48
How do you prove that the integral diverges?
– gimusi
Jul 27 at 6:48
I fixed it. I hope it's right
– J.Dane
Jul 27 at 7:03
I fixed it. I hope it's right
– J.Dane
Jul 27 at 7:03
It seems uncorrect, try to derive to check. Cauchy condensation test would simplify a lot.
– gimusi
Jul 27 at 7:09
It seems uncorrect, try to derive to check. Cauchy condensation test would simplify a lot.
– gimusi
Jul 27 at 7:09
What is $int u,mathrmdu$? It's not $frac u2+C$. Even when that's fixed, what about $alphane1$?
– robjohn♦
Jul 27 at 7:54
What is $int u,mathrmdu$? It's not $frac u2+C$. Even when that's fixed, what about $alphane1$?
– robjohn♦
Jul 27 at 7:54
Yes, I had my integral wrong.
– J.Dane
Jul 27 at 7:56
Yes, I had my integral wrong.
– J.Dane
Jul 27 at 7:56
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
If you know Bertrand's series, the answer is obvious by the comparison test:
A Bertrand's series is a series
$$sum_n=2^inftyfrac1n^alpha,log^beta nqquad (alpha,betainbf R)$$
and it is known to converge if and only if
- either $alpha>1$,
- or $alpha=1$ and $beta>1$.
answered Jul 27 at 8:46
Bernard
110k635102
So is this right $frac1n^alpha ltfrac1log^alphan $, $fraclog(logn)n^alpha ltfraclog(logn)log^alphan $, $frac1n^alphalog^-1(logn) ltfraclog(logn)log^alphan $ so $beta=-1$, therefore, the sequence diverges for $alpha lt 1$
– J.Dane
Jul 27 at 9:16
It's much simpler than that: $;fracloglog nlog^alpha n>frac1log^alpha n$ ($n>2$), and the latter series diverges.
– Bernard
Jul 27 at 9:16
yes for $alpha lt 1$
– J.Dane
Jul 27 at 9:17
@J.Dane No, for every $alpha$.
– Did
Jul 27 at 11:48
add a comment |Â
up vote
2
down vote
We have that eventually (notably for $n>e^e$)
$$fraclog(logn)log^alphange frac1log^alphan$$
and for $alpha>0$
$$sum_n=2^inftyfrac1log^alphan$$
diverges by limit comparison test with $sum_n=2^infty frac1n$ indeed
$$fracfrac1log^alphanfrac1n=fracnlog^alphanto infty$$
answered Jul 27 at 13:49
gimusi
65k73583
add a comment |Â
up vote
1
down vote
For all $xgt0$, $log(x)lt x$. Therefore, for all $xgt1$ and $alphagt0$,
$$
beginalign
log(x)^alpha
&=alpha^alphalog!left(x^1/alpharight)^alpha\
<alpha^alpha x^1/alphacdotalpha\[3pt]
&=alpha^alpha x
endalign
$$
Therefore, since $log(log(n))ge1$ for $nge16$, it suffices to compare
$$
sum_n=2^inftyfrac1log(n)^alphagefrac1alpha^alphasum_n=2^inftyfrac1n
$$
answered Jul 27 at 7:10
robjohn♦
258k25297612
Very nice and elegant!
– gimusi
Jul 27 at 8:01
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If you know Bertrand's series, the answer is obvious by the comparison test:
A Bertrand's series is a series
$$sum_n=2^inftyfrac1n^alpha,log^beta nqquad (alpha,betainbf R)$$
and it is known to converge if and only if
- either $alpha>1$,
- or $alpha=1$ and $beta>1$.
answered Jul 27 at 8:46
Bernard
110k635102
So is this right $frac1n^alpha ltfrac1log^alphan $, $fraclog(logn)n^alpha ltfraclog(logn)log^alphan $, $frac1n^alphalog^-1(logn) ltfraclog(logn)log^alphan $ so $beta=-1$, therefore, the sequence diverges for $alpha lt 1$
– J.Dane
Jul 27 at 9:16
It's much simpler than that: $;fracloglog nlog^alpha n>frac1log^alpha n$ ($n>2$), and the latter series diverges.
– Bernard
Jul 27 at 9:16
yes for $alpha lt 1$
– J.Dane
Jul 27 at 9:17
@J.Dane No, for every $alpha$.
– Did
Jul 27 at 11:48
add a comment |Â
up vote
1
down vote
accepted
If you know Bertrand's series, the answer is obvious by the comparison test:
A Bertrand's series is a series
$$sum_n=2^inftyfrac1n^alpha,log^beta nqquad (alpha,betainbf R)$$
and it is known to converge if and only if
- either $alpha>1$,
- or $alpha=1$ and $beta>1$.
answered Jul 27 at 8:46
Bernard
110k635102
So is this right $frac1n^alpha ltfrac1log^alphan $, $fraclog(logn)n^alpha ltfraclog(logn)log^alphan $, $frac1n^alphalog^-1(logn) ltfraclog(logn)log^alphan $ so $beta=-1$, therefore, the sequence diverges for $alpha lt 1$
– J.Dane
Jul 27 at 9:16
It's much simpler than that: $;fracloglog nlog^alpha n>frac1log^alpha n$ ($n>2$), and the latter series diverges.
– Bernard
Jul 27 at 9:16
yes for $alpha lt 1$
– J.Dane
Jul 27 at 9:17
@J.Dane No, for every $alpha$.
– Did
Jul 27 at 11:48
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If you know Bertrand's series, the answer is obvious by the comparison test:
A Bertrand's series is a series
$$sum_n=2^inftyfrac1n^alpha,log^beta nqquad (alpha,betainbf R)$$
and it is known to converge if and only if
- either $alpha>1$,
- or $alpha=1$ and $beta>1$.
answered Jul 27 at 8:46
Bernard
110k635102
If you know Bertrand's series, the answer is obvious by the comparison test:
A Bertrand's series is a series
$$sum_n=2^inftyfrac1n^alpha,log^beta nqquad (alpha,betainbf R)$$
and it is known to converge if and only if
- either $alpha>1$,
- or $alpha=1$ and $beta>1$.
answered Jul 27 at 8:46
Bernard
110k635102
answered Jul 27 at 8:46
Bernard
110k635102
answered Jul 27 at 8:46
Bernard
110k635102
answered Jul 27 at 8:46
Bernard
110k635102
110k635102
So is this right $frac1n^alpha ltfrac1log^alphan $, $fraclog(logn)n^alpha ltfraclog(logn)log^alphan $, $frac1n^alphalog^-1(logn) ltfraclog(logn)log^alphan $ so $beta=-1$, therefore, the sequence diverges for $alpha lt 1$
– J.Dane
Jul 27 at 9:16
It's much simpler than that: $;fracloglog nlog^alpha n>frac1log^alpha n$ ($n>2$), and the latter series diverges.
– Bernard
Jul 27 at 9:16
yes for $alpha lt 1$
– J.Dane
Jul 27 at 9:17
@J.Dane No, for every $alpha$.
– Did
Jul 27 at 11:48
add a comment |Â
So is this right $frac1n^alpha ltfrac1log^alphan $, $fraclog(logn)n^alpha ltfraclog(logn)log^alphan $, $frac1n^alphalog^-1(logn) ltfraclog(logn)log^alphan $ so $beta=-1$, therefore, the sequence diverges for $alpha lt 1$
– J.Dane
Jul 27 at 9:16
It's much simpler than that: $;fracloglog nlog^alpha n>frac1log^alpha n$ ($n>2$), and the latter series diverges.
– Bernard
Jul 27 at 9:16
yes for $alpha lt 1$
– J.Dane
Jul 27 at 9:17
@J.Dane No, for every $alpha$.
– Did
Jul 27 at 11:48
So is this right $frac1n^alpha ltfrac1log^alphan $, $fraclog(logn)n^alpha ltfraclog(logn)log^alphan $, $frac1n^alphalog^-1(logn) ltfraclog(logn)log^alphan $ so $beta=-1$, therefore, the sequence diverges for $alpha lt 1$
– J.Dane
Jul 27 at 9:16
So is this right $frac1n^alpha ltfrac1log^alphan $, $fraclog(logn)n^alpha ltfraclog(logn)log^alphan $, $frac1n^alphalog^-1(logn) ltfraclog(logn)log^alphan $ so $beta=-1$, therefore, the sequence diverges for $alpha lt 1$
– J.Dane
Jul 27 at 9:16
It's much simpler than that: $;fracloglog nlog^alpha n>frac1log^alpha n$ ($n>2$), and the latter series diverges.
– Bernard
Jul 27 at 9:16
It's much simpler than that: $;fracloglog nlog^alpha n>frac1log^alpha n$ ($n>2$), and the latter series diverges.
– Bernard
Jul 27 at 9:16
yes for $alpha lt 1$
– J.Dane
Jul 27 at 9:17
yes for $alpha lt 1$
– J.Dane
Jul 27 at 9:17
@J.Dane No, for every $alpha$.
– Did
Jul 27 at 11:48
@J.Dane No, for every $alpha$.
– Did
Jul 27 at 11:48
add a comment |Â
up vote
2
down vote
We have that eventually (notably for $n>e^e$)
$$fraclog(logn)log^alphange frac1log^alphan$$
and for $alpha>0$
$$sum_n=2^inftyfrac1log^alphan$$
diverges by limit comparison test with $sum_n=2^infty frac1n$ indeed
$$fracfrac1log^alphanfrac1n=fracnlog^alphanto infty$$
answered Jul 27 at 13:49
gimusi
65k73583
add a comment |Â
up vote
2
down vote
We have that eventually (notably for $n>e^e$)
$$fraclog(logn)log^alphange frac1log^alphan$$
and for $alpha>0$
$$sum_n=2^inftyfrac1log^alphan$$
diverges by limit comparison test with $sum_n=2^infty frac1n$ indeed
$$fracfrac1log^alphanfrac1n=fracnlog^alphanto infty$$
answered Jul 27 at 13:49
gimusi
65k73583
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We have that eventually (notably for $n>e^e$)
$$fraclog(logn)log^alphange frac1log^alphan$$
and for $alpha>0$
$$sum_n=2^inftyfrac1log^alphan$$
diverges by limit comparison test with $sum_n=2^infty frac1n$ indeed
$$fracfrac1log^alphanfrac1n=fracnlog^alphanto infty$$
answered Jul 27 at 13:49
gimusi
65k73583
We have that eventually (notably for $n>e^e$)
$$fraclog(logn)log^alphange frac1log^alphan$$
and for $alpha>0$
$$sum_n=2^inftyfrac1log^alphan$$
diverges by limit comparison test with $sum_n=2^infty frac1n$ indeed
$$fracfrac1log^alphanfrac1n=fracnlog^alphanto infty$$
answered Jul 27 at 13:49
gimusi
65k73583
edited Jul 27 at 14:21
answered Jul 27 at 13:49
gimusi
65k73583
answered Jul 27 at 13:49
gimusi
65k73583
answered Jul 27 at 13:49
gimusi
65k73583
65k73583
add a comment |Â
add a comment |Â
up vote
1
down vote
For all $xgt0$, $log(x)lt x$. Therefore, for all $xgt1$ and $alphagt0$,
$$
beginalign
log(x)^alpha
&=alpha^alphalog!left(x^1/alpharight)^alpha\
<alpha^alpha x^1/alphacdotalpha\[3pt]
&=alpha^alpha x
endalign
$$
Therefore, since $log(log(n))ge1$ for $nge16$, it suffices to compare
$$
sum_n=2^inftyfrac1log(n)^alphagefrac1alpha^alphasum_n=2^inftyfrac1n
$$
answered Jul 27 at 7:10
robjohn♦
258k25297612
Very nice and elegant!
– gimusi
Jul 27 at 8:01
add a comment |Â
up vote
1
down vote
For all $xgt0$, $log(x)lt x$. Therefore, for all $xgt1$ and $alphagt0$,
$$
beginalign
log(x)^alpha
&=alpha^alphalog!left(x^1/alpharight)^alpha\
<alpha^alpha x^1/alphacdotalpha\[3pt]
&=alpha^alpha x
endalign
$$
Therefore, since $log(log(n))ge1$ for $nge16$, it suffices to compare
$$
sum_n=2^inftyfrac1log(n)^alphagefrac1alpha^alphasum_n=2^inftyfrac1n
$$
answered Jul 27 at 7:10
robjohn♦
258k25297612
Very nice and elegant!
– gimusi
Jul 27 at 8:01
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For all $xgt0$, $log(x)lt x$. Therefore, for all $xgt1$ and $alphagt0$,
$$
beginalign
log(x)^alpha
&=alpha^alphalog!left(x^1/alpharight)^alpha\
<alpha^alpha x^1/alphacdotalpha\[3pt]
&=alpha^alpha x
endalign
$$
Therefore, since $log(log(n))ge1$ for $nge16$, it suffices to compare
$$
sum_n=2^inftyfrac1log(n)^alphagefrac1alpha^alphasum_n=2^inftyfrac1n
$$
answered Jul 27 at 7:10
robjohn♦
258k25297612
For all $xgt0$, $log(x)lt x$. Therefore, for all $xgt1$ and $alphagt0$,
$$
beginalign
log(x)^alpha
&=alpha^alphalog!left(x^1/alpharight)^alpha\
<alpha^alpha x^1/alphacdotalpha\[3pt]
&=alpha^alpha x
endalign
$$
Therefore, since $log(log(n))ge1$ for $nge16$, it suffices to compare
$$
sum_n=2^inftyfrac1log(n)^alphagefrac1alpha^alphasum_n=2^inftyfrac1n
$$
answered Jul 27 at 7:10
robjohn♦
258k25297612
answered Jul 27 at 7:10
robjohn♦
258k25297612
answered Jul 27 at 7:10
robjohn♦
258k25297612
answered Jul 27 at 7:10
robjohn♦
258k25297612
258k25297612
Very nice and elegant!
– gimusi
Jul 27 at 8:01
add a comment |Â
Very nice and elegant!
– gimusi
Jul 27 at 8:01
Very nice and elegant!
– gimusi
Jul 27 at 8:01
Very nice and elegant!
– gimusi
Jul 27 at 8:01
add a comment |Â
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How do you prove that the integral diverges?
– gimusi
Jul 27 at 6:48
I fixed it. I hope it's right
– J.Dane
Jul 27 at 7:03
It seems uncorrect, try to derive to check. Cauchy condensation test would simplify a lot.
– gimusi
Jul 27 at 7:09
What is $int u,mathrmdu$? It's not $frac u2+C$. Even when that's fixed, what about $alphane1$?
– robjohn♦
Jul 27 at 7:54
Yes, I had my integral wrong.
– J.Dane
Jul 27 at 7:56