Mixing
A test for convergence involving logarithms
$begingroup$
$$sum_{n=2}^infty frac{n^{log n}}{(log n)^n}$$
Tried to use the inequality of $ln x< x-1$ and apply the comparison theorem but no success.
calculus sequences-and-series
edited Jan 30 at 3:03
David G. Stork
11.6k41534
asked Jan 30 at 1:52
Nick KaklNick Kakl
176
$endgroup$
add a comment |
$begingroup$
$$sum_{n=2}^infty frac{n^{log n}}{(log n)^n}$$
Tried to use the inequality of $ln x< x-1$ and apply the comparison theorem but no success.
calculus sequences-and-series
edited Jan 30 at 3:03
David G. Stork
11.6k41534
asked Jan 30 at 1:52
Nick KaklNick Kakl
176
$endgroup$
$begingroup$
As tip for future posts, you can enclose exponents in braces{ }to capture the entire argument. That is,$n^{log n}$produces $n^{log n}$.
$endgroup$
– Clayton
Jan 30 at 1:54
add a comment |
$begingroup$
$$sum_{n=2}^infty frac{n^{log n}}{(log n)^n}$$
Tried to use the inequality of $ln x< x-1$ and apply the comparison theorem but no success.
calculus sequences-and-series
edited Jan 30 at 3:03
David G. Stork
11.6k41534
asked Jan 30 at 1:52
Nick KaklNick Kakl
176
$endgroup$
$$sum_{n=2}^infty frac{n^{log n}}{(log n)^n}$$
Tried to use the inequality of $ln x< x-1$ and apply the comparison theorem but no success.
calculus sequences-and-series
calculus sequences-and-series
edited Jan 30 at 3:03
David G. Stork
11.6k41534
asked Jan 30 at 1:52
Nick KaklNick Kakl
176
edited Jan 30 at 3:03
David G. Stork
11.6k41534
asked Jan 30 at 1:52
Nick KaklNick Kakl
176
edited Jan 30 at 3:03
David G. Stork
11.6k41534
edited Jan 30 at 3:03
David G. Stork
11.6k41534
edited Jan 30 at 3:03
David G. Stork
11.6k41534
11.6k41534
asked Jan 30 at 1:52
Nick KaklNick Kakl
176
asked Jan 30 at 1:52
Nick KaklNick Kakl
176
asked Jan 30 at 1:52
Nick KaklNick Kakl
176
176
$begingroup$
As tip for future posts, you can enclose exponents in braces{ }to capture the entire argument. That is,$n^{log n}$produces $n^{log n}$.
$endgroup$
– Clayton
Jan 30 at 1:54
add a comment |
$begingroup$
As tip for future posts, you can enclose exponents in braces{ }to capture the entire argument. That is,$n^{log n}$produces $n^{log n}$.
$endgroup$
– Clayton
Jan 30 at 1:54
$begingroup$
As tip for future posts, you can enclose exponents in braces
{ } to capture the entire argument. That is, $n^{log n}$ produces $n^{log n}$.$endgroup$
– Clayton
Jan 30 at 1:54
$begingroup$
As tip for future posts, you can enclose exponents in braces
{ } to capture the entire argument. That is, $n^{log n}$ produces $n^{log n}$.$endgroup$
– Clayton
Jan 30 at 1:54
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We consider
$$lim_{nto infty}sqrt[n]{frac{n^{log n}}{(log n)^n}}
=lim_{nto infty}frac{n^{frac{log n}{n}}}{log n}
=lim_{nto infty}frac{e^{frac{(log n)^2}{n}}}{log n}=0.$$
By Cauchy test, this series is convergent.
answered Jan 30 at 2:54
RiemannRiemann
3,5251422
$endgroup$
$begingroup$
Thnx for the answer, you got the limit with DLH?
$endgroup$
– Nick Kakl
Jan 30 at 2:59
$begingroup$
Yse, it is easy to get $frac{(log n)^2}{n}to 0$ as $nto infty$.
$endgroup$
– Riemann
Jan 30 at 3:03
add a comment |
$begingroup$
Let's upper-bound each term with a term whose sum converges. Note that, for $n$ large enough, $log log n geq 1$, therefore:
$$frac{n^{log n}}{(log n)^n} = frac{e^{log^2n}}{e^{nloglog n}}leq frac{e^{log^2n}}{e^{n}}=e^{-frac n 2}e^{-frac n 2+log^2n}$$
and because for $n$ large enough, $left(-frac n 2+log^2nright)leq 0$ (left as an exercise),
we must have $$frac{n^{log n}}{(log n)^n} leq e^{-frac n 2}$$
Conclusion: The sum converges
answered Jan 30 at 2:49
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
Ty for the answer :)
$endgroup$
– Nick Kakl
Jan 30 at 2:59
add a comment |
$begingroup$
Too long for a comment.
The ratio test is interesting to work
$$a_n=frac{n^{log n}}{(log n)^n}implies log(a_n)=log ^2(n)-n log (n)$$ Now, using Taylor for infinitely large values of $n$,
$$log(a_{n+1})-log(a_n)=-1-log left({n}right)+frac{2 log
left({n}right)-frac{1}{2}}{n}+Oleft(frac{1}{n^2}right)$$ and, continuing with Taylor expansions,
$$frac{a_{n+1} } {a_n }=e^{log(a_{n+1})-log(a_n) }=frac1 {e n}+frac{2 log
left({n}right)-frac{1}{2}}{en^2}+Oleft(frac{1}{n^3}right)$$
answered Jan 30 at 5:45
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We consider
$$lim_{nto infty}sqrt[n]{frac{n^{log n}}{(log n)^n}}
=lim_{nto infty}frac{n^{frac{log n}{n}}}{log n}
=lim_{nto infty}frac{e^{frac{(log n)^2}{n}}}{log n}=0.$$
By Cauchy test, this series is convergent.
answered Jan 30 at 2:54
RiemannRiemann
3,5251422
$endgroup$
$begingroup$
Thnx for the answer, you got the limit with DLH?
$endgroup$
– Nick Kakl
Jan 30 at 2:59
$begingroup$
Yse, it is easy to get $frac{(log n)^2}{n}to 0$ as $nto infty$.
$endgroup$
– Riemann
Jan 30 at 3:03
add a comment |
$begingroup$
We consider
$$lim_{nto infty}sqrt[n]{frac{n^{log n}}{(log n)^n}}
=lim_{nto infty}frac{n^{frac{log n}{n}}}{log n}
=lim_{nto infty}frac{e^{frac{(log n)^2}{n}}}{log n}=0.$$
By Cauchy test, this series is convergent.
answered Jan 30 at 2:54
RiemannRiemann
3,5251422
$endgroup$
$begingroup$
Thnx for the answer, you got the limit with DLH?
$endgroup$
– Nick Kakl
Jan 30 at 2:59
$begingroup$
Yse, it is easy to get $frac{(log n)^2}{n}to 0$ as $nto infty$.
$endgroup$
– Riemann
Jan 30 at 3:03
add a comment |
$begingroup$
We consider
$$lim_{nto infty}sqrt[n]{frac{n^{log n}}{(log n)^n}}
=lim_{nto infty}frac{n^{frac{log n}{n}}}{log n}
=lim_{nto infty}frac{e^{frac{(log n)^2}{n}}}{log n}=0.$$
By Cauchy test, this series is convergent.
answered Jan 30 at 2:54
RiemannRiemann
3,5251422
$endgroup$
We consider
$$lim_{nto infty}sqrt[n]{frac{n^{log n}}{(log n)^n}}
=lim_{nto infty}frac{n^{frac{log n}{n}}}{log n}
=lim_{nto infty}frac{e^{frac{(log n)^2}{n}}}{log n}=0.$$
By Cauchy test, this series is convergent.
answered Jan 30 at 2:54
RiemannRiemann
3,5251422
answered Jan 30 at 2:54
RiemannRiemann
3,5251422
answered Jan 30 at 2:54
RiemannRiemann
3,5251422
answered Jan 30 at 2:54
RiemannRiemann
3,5251422
3,5251422
$begingroup$
Thnx for the answer, you got the limit with DLH?
$endgroup$
– Nick Kakl
Jan 30 at 2:59
$begingroup$
Yse, it is easy to get $frac{(log n)^2}{n}to 0$ as $nto infty$.
$endgroup$
– Riemann
Jan 30 at 3:03
add a comment |
$begingroup$
Thnx for the answer, you got the limit with DLH?
$endgroup$
– Nick Kakl
Jan 30 at 2:59
$begingroup$
Yse, it is easy to get $frac{(log n)^2}{n}to 0$ as $nto infty$.
$endgroup$
– Riemann
Jan 30 at 3:03
$begingroup$
Thnx for the answer, you got the limit with DLH?
$endgroup$
– Nick Kakl
Jan 30 at 2:59
$begingroup$
Thnx for the answer, you got the limit with DLH?
$endgroup$
– Nick Kakl
Jan 30 at 2:59
$begingroup$
Yse, it is easy to get $frac{(log n)^2}{n}to 0$ as $nto infty$.
$endgroup$
– Riemann
Jan 30 at 3:03
$begingroup$
Yse, it is easy to get $frac{(log n)^2}{n}to 0$ as $nto infty$.
$endgroup$
– Riemann
Jan 30 at 3:03
add a comment |
$begingroup$
Let's upper-bound each term with a term whose sum converges. Note that, for $n$ large enough, $log log n geq 1$, therefore:
$$frac{n^{log n}}{(log n)^n} = frac{e^{log^2n}}{e^{nloglog n}}leq frac{e^{log^2n}}{e^{n}}=e^{-frac n 2}e^{-frac n 2+log^2n}$$
and because for $n$ large enough, $left(-frac n 2+log^2nright)leq 0$ (left as an exercise),
we must have $$frac{n^{log n}}{(log n)^n} leq e^{-frac n 2}$$
Conclusion: The sum converges
answered Jan 30 at 2:49
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
Ty for the answer :)
$endgroup$
– Nick Kakl
Jan 30 at 2:59
add a comment |
$begingroup$
Let's upper-bound each term with a term whose sum converges. Note that, for $n$ large enough, $log log n geq 1$, therefore:
$$frac{n^{log n}}{(log n)^n} = frac{e^{log^2n}}{e^{nloglog n}}leq frac{e^{log^2n}}{e^{n}}=e^{-frac n 2}e^{-frac n 2+log^2n}$$
and because for $n$ large enough, $left(-frac n 2+log^2nright)leq 0$ (left as an exercise),
we must have $$frac{n^{log n}}{(log n)^n} leq e^{-frac n 2}$$
Conclusion: The sum converges
answered Jan 30 at 2:49
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
Ty for the answer :)
$endgroup$
– Nick Kakl
Jan 30 at 2:59
add a comment |
$begingroup$
Let's upper-bound each term with a term whose sum converges. Note that, for $n$ large enough, $log log n geq 1$, therefore:
$$frac{n^{log n}}{(log n)^n} = frac{e^{log^2n}}{e^{nloglog n}}leq frac{e^{log^2n}}{e^{n}}=e^{-frac n 2}e^{-frac n 2+log^2n}$$
and because for $n$ large enough, $left(-frac n 2+log^2nright)leq 0$ (left as an exercise),
we must have $$frac{n^{log n}}{(log n)^n} leq e^{-frac n 2}$$
Conclusion: The sum converges
answered Jan 30 at 2:49
Stefan LafonStefan Lafon
3,005212
$endgroup$
Let's upper-bound each term with a term whose sum converges. Note that, for $n$ large enough, $log log n geq 1$, therefore:
$$frac{n^{log n}}{(log n)^n} = frac{e^{log^2n}}{e^{nloglog n}}leq frac{e^{log^2n}}{e^{n}}=e^{-frac n 2}e^{-frac n 2+log^2n}$$
and because for $n$ large enough, $left(-frac n 2+log^2nright)leq 0$ (left as an exercise),
we must have $$frac{n^{log n}}{(log n)^n} leq e^{-frac n 2}$$
Conclusion: The sum converges
answered Jan 30 at 2:49
Stefan LafonStefan Lafon
3,005212
answered Jan 30 at 2:49
Stefan LafonStefan Lafon
3,005212
answered Jan 30 at 2:49
Stefan LafonStefan Lafon
3,005212
answered Jan 30 at 2:49
Stefan LafonStefan Lafon
3,005212
3,005212
$begingroup$
Ty for the answer :)
$endgroup$
– Nick Kakl
Jan 30 at 2:59
add a comment |
$begingroup$
Ty for the answer :)
$endgroup$
– Nick Kakl
Jan 30 at 2:59
$begingroup$
Ty for the answer :)
$endgroup$
– Nick Kakl
Jan 30 at 2:59
$begingroup$
Ty for the answer :)
$endgroup$
– Nick Kakl
Jan 30 at 2:59
add a comment |
$begingroup$
Too long for a comment.
The ratio test is interesting to work
$$a_n=frac{n^{log n}}{(log n)^n}implies log(a_n)=log ^2(n)-n log (n)$$ Now, using Taylor for infinitely large values of $n$,
$$log(a_{n+1})-log(a_n)=-1-log left({n}right)+frac{2 log
left({n}right)-frac{1}{2}}{n}+Oleft(frac{1}{n^2}right)$$ and, continuing with Taylor expansions,
$$frac{a_{n+1} } {a_n }=e^{log(a_{n+1})-log(a_n) }=frac1 {e n}+frac{2 log
left({n}right)-frac{1}{2}}{en^2}+Oleft(frac{1}{n^3}right)$$
answered Jan 30 at 5:45
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
add a comment |
$begingroup$
Too long for a comment.
The ratio test is interesting to work
$$a_n=frac{n^{log n}}{(log n)^n}implies log(a_n)=log ^2(n)-n log (n)$$ Now, using Taylor for infinitely large values of $n$,
$$log(a_{n+1})-log(a_n)=-1-log left({n}right)+frac{2 log
left({n}right)-frac{1}{2}}{n}+Oleft(frac{1}{n^2}right)$$ and, continuing with Taylor expansions,
$$frac{a_{n+1} } {a_n }=e^{log(a_{n+1})-log(a_n) }=frac1 {e n}+frac{2 log
left({n}right)-frac{1}{2}}{en^2}+Oleft(frac{1}{n^3}right)$$
answered Jan 30 at 5:45
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
add a comment |
$begingroup$
Too long for a comment.
The ratio test is interesting to work
$$a_n=frac{n^{log n}}{(log n)^n}implies log(a_n)=log ^2(n)-n log (n)$$ Now, using Taylor for infinitely large values of $n$,
$$log(a_{n+1})-log(a_n)=-1-log left({n}right)+frac{2 log
left({n}right)-frac{1}{2}}{n}+Oleft(frac{1}{n^2}right)$$ and, continuing with Taylor expansions,
$$frac{a_{n+1} } {a_n }=e^{log(a_{n+1})-log(a_n) }=frac1 {e n}+frac{2 log
left({n}right)-frac{1}{2}}{en^2}+Oleft(frac{1}{n^3}right)$$
answered Jan 30 at 5:45
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
Too long for a comment.
The ratio test is interesting to work
$$a_n=frac{n^{log n}}{(log n)^n}implies log(a_n)=log ^2(n)-n log (n)$$ Now, using Taylor for infinitely large values of $n$,
$$log(a_{n+1})-log(a_n)=-1-log left({n}right)+frac{2 log
left({n}right)-frac{1}{2}}{n}+Oleft(frac{1}{n^2}right)$$ and, continuing with Taylor expansions,
$$frac{a_{n+1} } {a_n }=e^{log(a_{n+1})-log(a_n) }=frac1 {e n}+frac{2 log
left({n}right)-frac{1}{2}}{en^2}+Oleft(frac{1}{n^3}right)$$
answered Jan 30 at 5:45
Claude LeiboviciClaude Leibovici
125k1158136
answered Jan 30 at 5:45
Claude LeiboviciClaude Leibovici
125k1158136
answered Jan 30 at 5:45
Claude LeiboviciClaude Leibovici
125k1158136
answered Jan 30 at 5:45
Claude LeiboviciClaude Leibovici
125k1158136
125k1158136
add a comment |
add a comment |
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$begingroup$
As tip for future posts, you can enclose exponents in braces
{ }to capture the entire argument. That is,$n^{log n}$produces $n^{log n}$.$endgroup$
– Clayton
Jan 30 at 1:54