Mixing
Convergence of the series $sum_{n=1}^{infty} frac{(-1)^n}{n^{left(1+frac{1}{ln(ln(n))}right)}}$
$begingroup$
Study the convergence of the following series:
$$sum_{n=1}^{infty} frac{(-1)^n}{n^{left(1+frac{1}{ln(ln(n))}right)}}$$
The absolute convergence is giving me hard times, eventually I was following this path, any tips?
$left( n^{left(1+frac{1}{ln(ln(n))}right)}right)^{-1} = left( e^{ln(n)+frac{1}{ln(ln(n))}ln(n)}right)^{-1} = left( e^{ln(n)left(1+frac{1}{ln(ln(n))}right)}right)^{-1} =,,,,$?
real-analysis sequences-and-series convergence
asked Jan 21 at 21:11
F.incF.inc
315110
$endgroup$
add a comment |
$begingroup$
Study the convergence of the following series:
$$sum_{n=1}^{infty} frac{(-1)^n}{n^{left(1+frac{1}{ln(ln(n))}right)}}$$
The absolute convergence is giving me hard times, eventually I was following this path, any tips?
$left( n^{left(1+frac{1}{ln(ln(n))}right)}right)^{-1} = left( e^{ln(n)+frac{1}{ln(ln(n))}ln(n)}right)^{-1} = left( e^{ln(n)left(1+frac{1}{ln(ln(n))}right)}right)^{-1} =,,,,$?
real-analysis sequences-and-series convergence
asked Jan 21 at 21:11
F.incF.inc
315110
$endgroup$
1
$begingroup$
Hint: $frac{ln{n}}{ln{ln{n}}} geq 2ln{ln{n}}$ when $n$ is large.
$endgroup$
– Mindlack
Jan 21 at 21:14
$begingroup$
@Mindlack how to show that fact? Thank you
$endgroup$
– F.inc
Jan 21 at 21:24
1
$begingroup$
If $x geq C >0$, $x geq 2(ln{x})^2$. So if $n geq e^C$, ...
$endgroup$
– Mindlack
Jan 21 at 21:26
add a comment |
$begingroup$
Study the convergence of the following series:
$$sum_{n=1}^{infty} frac{(-1)^n}{n^{left(1+frac{1}{ln(ln(n))}right)}}$$
The absolute convergence is giving me hard times, eventually I was following this path, any tips?
$left( n^{left(1+frac{1}{ln(ln(n))}right)}right)^{-1} = left( e^{ln(n)+frac{1}{ln(ln(n))}ln(n)}right)^{-1} = left( e^{ln(n)left(1+frac{1}{ln(ln(n))}right)}right)^{-1} =,,,,$?
real-analysis sequences-and-series convergence
asked Jan 21 at 21:11
F.incF.inc
315110
$endgroup$
Study the convergence of the following series:
$$sum_{n=1}^{infty} frac{(-1)^n}{n^{left(1+frac{1}{ln(ln(n))}right)}}$$
The absolute convergence is giving me hard times, eventually I was following this path, any tips?
$left( n^{left(1+frac{1}{ln(ln(n))}right)}right)^{-1} = left( e^{ln(n)+frac{1}{ln(ln(n))}ln(n)}right)^{-1} = left( e^{ln(n)left(1+frac{1}{ln(ln(n))}right)}right)^{-1} =,,,,$?
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
asked Jan 21 at 21:11
F.incF.inc
315110
asked Jan 21 at 21:11
F.incF.inc
315110
asked Jan 21 at 21:11
F.incF.inc
315110
asked Jan 21 at 21:11
F.incF.inc
315110
asked Jan 21 at 21:11
F.incF.inc
315110
315110
1
$begingroup$
Hint: $frac{ln{n}}{ln{ln{n}}} geq 2ln{ln{n}}$ when $n$ is large.
$endgroup$
– Mindlack
Jan 21 at 21:14
$begingroup$
@Mindlack how to show that fact? Thank you
$endgroup$
– F.inc
Jan 21 at 21:24
1
$begingroup$
If $x geq C >0$, $x geq 2(ln{x})^2$. So if $n geq e^C$, ...
$endgroup$
– Mindlack
Jan 21 at 21:26
add a comment |
1
$begingroup$
Hint: $frac{ln{n}}{ln{ln{n}}} geq 2ln{ln{n}}$ when $n$ is large.
$endgroup$
– Mindlack
Jan 21 at 21:14
$begingroup$
@Mindlack how to show that fact? Thank you
$endgroup$
– F.inc
Jan 21 at 21:24
1
$begingroup$
If $x geq C >0$, $x geq 2(ln{x})^2$. So if $n geq e^C$, ...
$endgroup$
– Mindlack
Jan 21 at 21:26
1
1
$begingroup$
Hint: $frac{ln{n}}{ln{ln{n}}} geq 2ln{ln{n}}$ when $n$ is large.
$endgroup$
– Mindlack
Jan 21 at 21:14
$begingroup$
Hint: $frac{ln{n}}{ln{ln{n}}} geq 2ln{ln{n}}$ when $n$ is large.
$endgroup$
– Mindlack
Jan 21 at 21:14
$begingroup$
@Mindlack how to show that fact? Thank you
$endgroup$
– F.inc
Jan 21 at 21:24
$begingroup$
@Mindlack how to show that fact? Thank you
$endgroup$
– F.inc
Jan 21 at 21:24
1
1
$begingroup$
If $x geq C >0$, $x geq 2(ln{x})^2$. So if $n geq e^C$, ...
$endgroup$
– Mindlack
Jan 21 at 21:26
$begingroup$
If $x geq C >0$, $x geq 2(ln{x})^2$. So if $n geq e^C$, ...
$endgroup$
– Mindlack
Jan 21 at 21:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since
$dfrac{x}{ln(x)}
to infty
$
as
$x to infty$,
it follows that
$dfrac{f(x)}{f(ln(x))}
to infty
$
as
$x to infty$
where $f$ is any positive,
strictly increasing,
continuous,
unbounded function.
Take
$f(x) = ln(x)$
to get the desired result.
answered Jan 21 at 23:12
marty cohenmarty cohen
74.2k549128
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since
$dfrac{x}{ln(x)}
to infty
$
as
$x to infty$,
it follows that
$dfrac{f(x)}{f(ln(x))}
to infty
$
as
$x to infty$
where $f$ is any positive,
strictly increasing,
continuous,
unbounded function.
Take
$f(x) = ln(x)$
to get the desired result.
answered Jan 21 at 23:12
marty cohenmarty cohen
74.2k549128
$endgroup$
add a comment |
$begingroup$
Since
$dfrac{x}{ln(x)}
to infty
$
as
$x to infty$,
it follows that
$dfrac{f(x)}{f(ln(x))}
to infty
$
as
$x to infty$
where $f$ is any positive,
strictly increasing,
continuous,
unbounded function.
Take
$f(x) = ln(x)$
to get the desired result.
answered Jan 21 at 23:12
marty cohenmarty cohen
74.2k549128
$endgroup$
add a comment |
$begingroup$
Since
$dfrac{x}{ln(x)}
to infty
$
as
$x to infty$,
it follows that
$dfrac{f(x)}{f(ln(x))}
to infty
$
as
$x to infty$
where $f$ is any positive,
strictly increasing,
continuous,
unbounded function.
Take
$f(x) = ln(x)$
to get the desired result.
answered Jan 21 at 23:12
marty cohenmarty cohen
74.2k549128
$endgroup$
Since
$dfrac{x}{ln(x)}
to infty
$
as
$x to infty$,
it follows that
$dfrac{f(x)}{f(ln(x))}
to infty
$
as
$x to infty$
where $f$ is any positive,
strictly increasing,
continuous,
unbounded function.
Take
$f(x) = ln(x)$
to get the desired result.
answered Jan 21 at 23:12
marty cohenmarty cohen
74.2k549128
answered Jan 21 at 23:12
marty cohenmarty cohen
74.2k549128
answered Jan 21 at 23:12
marty cohenmarty cohen
74.2k549128
answered Jan 21 at 23:12
marty cohenmarty cohen
74.2k549128
74.2k549128
add a comment |
add a comment |
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1
$begingroup$
Hint: $frac{ln{n}}{ln{ln{n}}} geq 2ln{ln{n}}$ when $n$ is large.
$endgroup$
– Mindlack
Jan 21 at 21:14
$begingroup$
@Mindlack how to show that fact? Thank you
$endgroup$
– F.inc
Jan 21 at 21:24
1
$begingroup$
If $x geq C >0$, $x geq 2(ln{x})^2$. So if $n geq e^C$, ...
$endgroup$
– Mindlack
Jan 21 at 21:26