how to show $ sum_{10}^{infty} frac{sin{frac{1}{n}}}{ln(n)}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})$...
$$ sum_{10}^{infty} frac{sin{frac{1}{n}}}{ln(n)}left(e^{frac{1}{n^2}} - 1right)left(sqrt{n^4 - 8}right) $$
I have tried a lot of stuff, didn't work at all . A hint will be good too.
I know that $sin(frac{1}{n}) < frac{1}{n}$. I have tried to show with Cauchy that it diverges.
sequences-and-series convergence
add a comment |
$$ sum_{10}^{infty} frac{sin{frac{1}{n}}}{ln(n)}left(e^{frac{1}{n^2}} - 1right)left(sqrt{n^4 - 8}right) $$
I have tried a lot of stuff, didn't work at all . A hint will be good too.
I know that $sin(frac{1}{n}) < frac{1}{n}$. I have tried to show with Cauchy that it diverges.
sequences-and-series convergence
4
limit comparison?
– Lord Shark the Unknown
2 days ago
yea you're gonna need to use comparison test
– user29418
2 days ago
the given series diverges
– Dr. Sonnhard Graubner
2 days ago
add a comment |
$$ sum_{10}^{infty} frac{sin{frac{1}{n}}}{ln(n)}left(e^{frac{1}{n^2}} - 1right)left(sqrt{n^4 - 8}right) $$
I have tried a lot of stuff, didn't work at all . A hint will be good too.
I know that $sin(frac{1}{n}) < frac{1}{n}$. I have tried to show with Cauchy that it diverges.
sequences-and-series convergence
$$ sum_{10}^{infty} frac{sin{frac{1}{n}}}{ln(n)}left(e^{frac{1}{n^2}} - 1right)left(sqrt{n^4 - 8}right) $$
I have tried a lot of stuff, didn't work at all . A hint will be good too.
I know that $sin(frac{1}{n}) < frac{1}{n}$. I have tried to show with Cauchy that it diverges.
sequences-and-series convergence
sequences-and-series convergence
edited 2 days ago
Dando18
4,66741235
4,66741235
asked 2 days ago
Mather
1517
1517
4
limit comparison?
– Lord Shark the Unknown
2 days ago
yea you're gonna need to use comparison test
– user29418
2 days ago
the given series diverges
– Dr. Sonnhard Graubner
2 days ago
add a comment |
4
limit comparison?
– Lord Shark the Unknown
2 days ago
yea you're gonna need to use comparison test
– user29418
2 days ago
the given series diverges
– Dr. Sonnhard Graubner
2 days ago
4
4
limit comparison?
– Lord Shark the Unknown
2 days ago
limit comparison?
– Lord Shark the Unknown
2 days ago
yea you're gonna need to use comparison test
– user29418
2 days ago
yea you're gonna need to use comparison test
– user29418
2 days ago
the given series diverges
– Dr. Sonnhard Graubner
2 days ago
the given series diverges
– Dr. Sonnhard Graubner
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
Because $sin(x)$ is concave on $left[0,fracpi2right]$, for $xinleft[0,fracpi2right]$,
$$
frac{sin(x)-sin(0)}{x-0}gefrac{sinleft(fracpi2right)-sin(0)}{fracpi2-0}tag1
$$
Therefore,
$$
sin(x)gefrac{2x}pitag2
$$
Since $e^x$ is convex,
$$
begin{align}
frac{e^x-e^0}{x-0}
&gelim_{tto0}frac{e^t-e^0}{t-0}\
&=e^0\[6pt]
&=1tag3
end{align}
$$
Therefore,
$$
e^xge1+xtag4
$$
and that
$$
begin{align}
x-sqrt{x^2-a}
&=frac{a}{x+sqrt{x^2-a}}\
&lefrac axtag5
end{align}
$$
Therefore,
$$
sqrt{x^2-a}ge x-frac axtag6
$$
Thus, applying $color{#C00}{(2)}$, $color{#090}{(4)}$, and $color{#00F}{(6)}$, for $nge2$, we have
$$
begin{align}
frac{color{#C00}{sinleft(frac1nright)}}{log(n)}color{#090}{left(e^{frac1{n^2}}-1right)}color{#00F}{sqrt{n^4-8}}
&gefrac{color{#C00}{frac2{pi n}}}{log(n)}color{#090}{left(frac1{n^2}right)}color{#00F}{left(n^2-frac8{n^2}right)}\
&=frac2pifrac1{nlog(n)}left(1-frac8{n^4}right)\
&gefrac1pifrac1{nlog(n)}tag7
end{align}
$$
The series $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ diverges by the Cauchy Condensation test (applied twice is nice); and therefore, the original series diverges by comparison with $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ using $(7)$.
add a comment |
For large $n$, your sequence behaves as
$$frac{1}{nlog n}$$
Because the following integral diverges,
$$int_{10}^infty frac{dx}{xlog x}=log log x Big|_{10}^infty $$
By the integral test, your series also diverges.
how did you know that it behaves like that
– Mather
2 days ago
Well $sin 1/n sim 1/n$, $sqrt{n^4-8}sim n^2$, and $e^{1/n^2}-1sim 1/n^2$. You can prove these asymptotics using the functions' Maclaurin series.
– Zachary
2 days ago
oh nice fast and subtle , i would have give this one as a very good answer too !
– Mather
2 days ago
thank you @Zachary
– Mather
2 days ago
No problem! @Mather
– Zachary
yesterday
add a comment |
Hint. One may observe that,
$$
lim_{ nto infty}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})= 1
$$giving, for a certain $n_0ge10$, $nge n_0$,
$$
frac12 le(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8}) tag1
$$ then, as $n ge 10$,
$$
frac{1}{n}-frac{1}{6n^3}lesin{frac{1}{n}}tag2
$$ yielding, for $Nge n_0$,
$$
frac{1}{2}sum_{n_0}^{N}frac{1}{nln n}-frac{1}{12}sum_{n_0}^{N}frac{1}{n^3ln n}lesum_{n_0}^{N} frac{sin{frac{1}{n}}}{ln n}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})
$$ leading to the divergence of the given series.
how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
– Mather
2 days ago
and for $(2)$ how did you get this inequality
– Mather
2 days ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Because $sin(x)$ is concave on $left[0,fracpi2right]$, for $xinleft[0,fracpi2right]$,
$$
frac{sin(x)-sin(0)}{x-0}gefrac{sinleft(fracpi2right)-sin(0)}{fracpi2-0}tag1
$$
Therefore,
$$
sin(x)gefrac{2x}pitag2
$$
Since $e^x$ is convex,
$$
begin{align}
frac{e^x-e^0}{x-0}
&gelim_{tto0}frac{e^t-e^0}{t-0}\
&=e^0\[6pt]
&=1tag3
end{align}
$$
Therefore,
$$
e^xge1+xtag4
$$
and that
$$
begin{align}
x-sqrt{x^2-a}
&=frac{a}{x+sqrt{x^2-a}}\
&lefrac axtag5
end{align}
$$
Therefore,
$$
sqrt{x^2-a}ge x-frac axtag6
$$
Thus, applying $color{#C00}{(2)}$, $color{#090}{(4)}$, and $color{#00F}{(6)}$, for $nge2$, we have
$$
begin{align}
frac{color{#C00}{sinleft(frac1nright)}}{log(n)}color{#090}{left(e^{frac1{n^2}}-1right)}color{#00F}{sqrt{n^4-8}}
&gefrac{color{#C00}{frac2{pi n}}}{log(n)}color{#090}{left(frac1{n^2}right)}color{#00F}{left(n^2-frac8{n^2}right)}\
&=frac2pifrac1{nlog(n)}left(1-frac8{n^4}right)\
&gefrac1pifrac1{nlog(n)}tag7
end{align}
$$
The series $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ diverges by the Cauchy Condensation test (applied twice is nice); and therefore, the original series diverges by comparison with $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ using $(7)$.
add a comment |
Because $sin(x)$ is concave on $left[0,fracpi2right]$, for $xinleft[0,fracpi2right]$,
$$
frac{sin(x)-sin(0)}{x-0}gefrac{sinleft(fracpi2right)-sin(0)}{fracpi2-0}tag1
$$
Therefore,
$$
sin(x)gefrac{2x}pitag2
$$
Since $e^x$ is convex,
$$
begin{align}
frac{e^x-e^0}{x-0}
&gelim_{tto0}frac{e^t-e^0}{t-0}\
&=e^0\[6pt]
&=1tag3
end{align}
$$
Therefore,
$$
e^xge1+xtag4
$$
and that
$$
begin{align}
x-sqrt{x^2-a}
&=frac{a}{x+sqrt{x^2-a}}\
&lefrac axtag5
end{align}
$$
Therefore,
$$
sqrt{x^2-a}ge x-frac axtag6
$$
Thus, applying $color{#C00}{(2)}$, $color{#090}{(4)}$, and $color{#00F}{(6)}$, for $nge2$, we have
$$
begin{align}
frac{color{#C00}{sinleft(frac1nright)}}{log(n)}color{#090}{left(e^{frac1{n^2}}-1right)}color{#00F}{sqrt{n^4-8}}
&gefrac{color{#C00}{frac2{pi n}}}{log(n)}color{#090}{left(frac1{n^2}right)}color{#00F}{left(n^2-frac8{n^2}right)}\
&=frac2pifrac1{nlog(n)}left(1-frac8{n^4}right)\
&gefrac1pifrac1{nlog(n)}tag7
end{align}
$$
The series $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ diverges by the Cauchy Condensation test (applied twice is nice); and therefore, the original series diverges by comparison with $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ using $(7)$.
add a comment |
Because $sin(x)$ is concave on $left[0,fracpi2right]$, for $xinleft[0,fracpi2right]$,
$$
frac{sin(x)-sin(0)}{x-0}gefrac{sinleft(fracpi2right)-sin(0)}{fracpi2-0}tag1
$$
Therefore,
$$
sin(x)gefrac{2x}pitag2
$$
Since $e^x$ is convex,
$$
begin{align}
frac{e^x-e^0}{x-0}
&gelim_{tto0}frac{e^t-e^0}{t-0}\
&=e^0\[6pt]
&=1tag3
end{align}
$$
Therefore,
$$
e^xge1+xtag4
$$
and that
$$
begin{align}
x-sqrt{x^2-a}
&=frac{a}{x+sqrt{x^2-a}}\
&lefrac axtag5
end{align}
$$
Therefore,
$$
sqrt{x^2-a}ge x-frac axtag6
$$
Thus, applying $color{#C00}{(2)}$, $color{#090}{(4)}$, and $color{#00F}{(6)}$, for $nge2$, we have
$$
begin{align}
frac{color{#C00}{sinleft(frac1nright)}}{log(n)}color{#090}{left(e^{frac1{n^2}}-1right)}color{#00F}{sqrt{n^4-8}}
&gefrac{color{#C00}{frac2{pi n}}}{log(n)}color{#090}{left(frac1{n^2}right)}color{#00F}{left(n^2-frac8{n^2}right)}\
&=frac2pifrac1{nlog(n)}left(1-frac8{n^4}right)\
&gefrac1pifrac1{nlog(n)}tag7
end{align}
$$
The series $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ diverges by the Cauchy Condensation test (applied twice is nice); and therefore, the original series diverges by comparison with $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ using $(7)$.
Because $sin(x)$ is concave on $left[0,fracpi2right]$, for $xinleft[0,fracpi2right]$,
$$
frac{sin(x)-sin(0)}{x-0}gefrac{sinleft(fracpi2right)-sin(0)}{fracpi2-0}tag1
$$
Therefore,
$$
sin(x)gefrac{2x}pitag2
$$
Since $e^x$ is convex,
$$
begin{align}
frac{e^x-e^0}{x-0}
&gelim_{tto0}frac{e^t-e^0}{t-0}\
&=e^0\[6pt]
&=1tag3
end{align}
$$
Therefore,
$$
e^xge1+xtag4
$$
and that
$$
begin{align}
x-sqrt{x^2-a}
&=frac{a}{x+sqrt{x^2-a}}\
&lefrac axtag5
end{align}
$$
Therefore,
$$
sqrt{x^2-a}ge x-frac axtag6
$$
Thus, applying $color{#C00}{(2)}$, $color{#090}{(4)}$, and $color{#00F}{(6)}$, for $nge2$, we have
$$
begin{align}
frac{color{#C00}{sinleft(frac1nright)}}{log(n)}color{#090}{left(e^{frac1{n^2}}-1right)}color{#00F}{sqrt{n^4-8}}
&gefrac{color{#C00}{frac2{pi n}}}{log(n)}color{#090}{left(frac1{n^2}right)}color{#00F}{left(n^2-frac8{n^2}right)}\
&=frac2pifrac1{nlog(n)}left(1-frac8{n^4}right)\
&gefrac1pifrac1{nlog(n)}tag7
end{align}
$$
The series $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ diverges by the Cauchy Condensation test (applied twice is nice); and therefore, the original series diverges by comparison with $sumlimits_{n=2}^inftyfrac1{nlog(n)}$ using $(7)$.
answered 2 days ago
robjohn♦
265k27303624
265k27303624
add a comment |
add a comment |
For large $n$, your sequence behaves as
$$frac{1}{nlog n}$$
Because the following integral diverges,
$$int_{10}^infty frac{dx}{xlog x}=log log x Big|_{10}^infty $$
By the integral test, your series also diverges.
how did you know that it behaves like that
– Mather
2 days ago
Well $sin 1/n sim 1/n$, $sqrt{n^4-8}sim n^2$, and $e^{1/n^2}-1sim 1/n^2$. You can prove these asymptotics using the functions' Maclaurin series.
– Zachary
2 days ago
oh nice fast and subtle , i would have give this one as a very good answer too !
– Mather
2 days ago
thank you @Zachary
– Mather
2 days ago
No problem! @Mather
– Zachary
yesterday
add a comment |
For large $n$, your sequence behaves as
$$frac{1}{nlog n}$$
Because the following integral diverges,
$$int_{10}^infty frac{dx}{xlog x}=log log x Big|_{10}^infty $$
By the integral test, your series also diverges.
how did you know that it behaves like that
– Mather
2 days ago
Well $sin 1/n sim 1/n$, $sqrt{n^4-8}sim n^2$, and $e^{1/n^2}-1sim 1/n^2$. You can prove these asymptotics using the functions' Maclaurin series.
– Zachary
2 days ago
oh nice fast and subtle , i would have give this one as a very good answer too !
– Mather
2 days ago
thank you @Zachary
– Mather
2 days ago
No problem! @Mather
– Zachary
yesterday
add a comment |
For large $n$, your sequence behaves as
$$frac{1}{nlog n}$$
Because the following integral diverges,
$$int_{10}^infty frac{dx}{xlog x}=log log x Big|_{10}^infty $$
By the integral test, your series also diverges.
For large $n$, your sequence behaves as
$$frac{1}{nlog n}$$
Because the following integral diverges,
$$int_{10}^infty frac{dx}{xlog x}=log log x Big|_{10}^infty $$
By the integral test, your series also diverges.
answered 2 days ago
Zachary
2,2971213
2,2971213
how did you know that it behaves like that
– Mather
2 days ago
Well $sin 1/n sim 1/n$, $sqrt{n^4-8}sim n^2$, and $e^{1/n^2}-1sim 1/n^2$. You can prove these asymptotics using the functions' Maclaurin series.
– Zachary
2 days ago
oh nice fast and subtle , i would have give this one as a very good answer too !
– Mather
2 days ago
thank you @Zachary
– Mather
2 days ago
No problem! @Mather
– Zachary
yesterday
add a comment |
how did you know that it behaves like that
– Mather
2 days ago
Well $sin 1/n sim 1/n$, $sqrt{n^4-8}sim n^2$, and $e^{1/n^2}-1sim 1/n^2$. You can prove these asymptotics using the functions' Maclaurin series.
– Zachary
2 days ago
oh nice fast and subtle , i would have give this one as a very good answer too !
– Mather
2 days ago
thank you @Zachary
– Mather
2 days ago
No problem! @Mather
– Zachary
yesterday
how did you know that it behaves like that
– Mather
2 days ago
how did you know that it behaves like that
– Mather
2 days ago
Well $sin 1/n sim 1/n$, $sqrt{n^4-8}sim n^2$, and $e^{1/n^2}-1sim 1/n^2$. You can prove these asymptotics using the functions' Maclaurin series.
– Zachary
2 days ago
Well $sin 1/n sim 1/n$, $sqrt{n^4-8}sim n^2$, and $e^{1/n^2}-1sim 1/n^2$. You can prove these asymptotics using the functions' Maclaurin series.
– Zachary
2 days ago
oh nice fast and subtle , i would have give this one as a very good answer too !
– Mather
2 days ago
oh nice fast and subtle , i would have give this one as a very good answer too !
– Mather
2 days ago
thank you @Zachary
– Mather
2 days ago
thank you @Zachary
– Mather
2 days ago
No problem! @Mather
– Zachary
yesterday
No problem! @Mather
– Zachary
yesterday
add a comment |
Hint. One may observe that,
$$
lim_{ nto infty}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})= 1
$$giving, for a certain $n_0ge10$, $nge n_0$,
$$
frac12 le(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8}) tag1
$$ then, as $n ge 10$,
$$
frac{1}{n}-frac{1}{6n^3}lesin{frac{1}{n}}tag2
$$ yielding, for $Nge n_0$,
$$
frac{1}{2}sum_{n_0}^{N}frac{1}{nln n}-frac{1}{12}sum_{n_0}^{N}frac{1}{n^3ln n}lesum_{n_0}^{N} frac{sin{frac{1}{n}}}{ln n}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})
$$ leading to the divergence of the given series.
how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
– Mather
2 days ago
and for $(2)$ how did you get this inequality
– Mather
2 days ago
add a comment |
Hint. One may observe that,
$$
lim_{ nto infty}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})= 1
$$giving, for a certain $n_0ge10$, $nge n_0$,
$$
frac12 le(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8}) tag1
$$ then, as $n ge 10$,
$$
frac{1}{n}-frac{1}{6n^3}lesin{frac{1}{n}}tag2
$$ yielding, for $Nge n_0$,
$$
frac{1}{2}sum_{n_0}^{N}frac{1}{nln n}-frac{1}{12}sum_{n_0}^{N}frac{1}{n^3ln n}lesum_{n_0}^{N} frac{sin{frac{1}{n}}}{ln n}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})
$$ leading to the divergence of the given series.
how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
– Mather
2 days ago
and for $(2)$ how did you get this inequality
– Mather
2 days ago
add a comment |
Hint. One may observe that,
$$
lim_{ nto infty}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})= 1
$$giving, for a certain $n_0ge10$, $nge n_0$,
$$
frac12 le(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8}) tag1
$$ then, as $n ge 10$,
$$
frac{1}{n}-frac{1}{6n^3}lesin{frac{1}{n}}tag2
$$ yielding, for $Nge n_0$,
$$
frac{1}{2}sum_{n_0}^{N}frac{1}{nln n}-frac{1}{12}sum_{n_0}^{N}frac{1}{n^3ln n}lesum_{n_0}^{N} frac{sin{frac{1}{n}}}{ln n}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})
$$ leading to the divergence of the given series.
Hint. One may observe that,
$$
lim_{ nto infty}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})= 1
$$giving, for a certain $n_0ge10$, $nge n_0$,
$$
frac12 le(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8}) tag1
$$ then, as $n ge 10$,
$$
frac{1}{n}-frac{1}{6n^3}lesin{frac{1}{n}}tag2
$$ yielding, for $Nge n_0$,
$$
frac{1}{2}sum_{n_0}^{N}frac{1}{nln n}-frac{1}{12}sum_{n_0}^{N}frac{1}{n^3ln n}lesum_{n_0}^{N} frac{sin{frac{1}{n}}}{ln n}(e^{frac{1}{n^2}} - 1)(sqrt{n^4 - 8})
$$ leading to the divergence of the given series.
edited 2 days ago
answered 2 days ago
Olivier Oloa
108k17176293
108k17176293
how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
– Mather
2 days ago
and for $(2)$ how did you get this inequality
– Mather
2 days ago
add a comment |
how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
– Mather
2 days ago
and for $(2)$ how did you get this inequality
– Mather
2 days ago
how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
– Mather
2 days ago
how did you know all of these facts ? say $(1)$ ? did you plug $epsilon$ = $ frac{1}{2} $ and get n ?
– Mather
2 days ago
and for $(2)$ how did you get this inequality
– Mather
2 days ago
and for $(2)$ how did you get this inequality
– Mather
2 days ago
add a comment |
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4
limit comparison?
– Lord Shark the Unknown
2 days ago
yea you're gonna need to use comparison test
– user29418
2 days ago
the given series diverges
– Dr. Sonnhard Graubner
2 days ago