Question on Convergent Series
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This is the last question on my real analysis final exam. I didn't get to spend much time on it. Hopefully I remember the problem correctly.
${a_n}_{ngeq1}$ is a sequence of real numbers such that $sum_{ngeq1}a_n$ converges. Let $b_n$ be the cardinality of the set ${kin N:a_k>frac{1}{2^n}}$. Show that $limsup_{nrightarrowinfty}frac{1}{2^n}b_n=0$.
My thought is since $sum frac{1}{2^n}$ converges and $sum a_n$ converges, their terms need to be "comparable", but I don't know how to capture this.
real-analysis sequences-and-series
edited Jan 19 at 3:58
Martin Sleziak
45.1k10123277
asked Dec 19 '18 at 11:11
Fluffy SkyeFluffy Skye
34019
$endgroup$
add a comment |
$begingroup$
This is the last question on my real analysis final exam. I didn't get to spend much time on it. Hopefully I remember the problem correctly.
${a_n}_{ngeq1}$ is a sequence of real numbers such that $sum_{ngeq1}a_n$ converges. Let $b_n$ be the cardinality of the set ${kin N:a_k>frac{1}{2^n}}$. Show that $limsup_{nrightarrowinfty}frac{1}{2^n}b_n=0$.
My thought is since $sum frac{1}{2^n}$ converges and $sum a_n$ converges, their terms need to be "comparable", but I don't know how to capture this.
real-analysis sequences-and-series
edited Jan 19 at 3:58
Martin Sleziak
45.1k10123277
asked Dec 19 '18 at 11:11
Fluffy SkyeFluffy Skye
34019
$endgroup$
add a comment |
$begingroup$
This is the last question on my real analysis final exam. I didn't get to spend much time on it. Hopefully I remember the problem correctly.
${a_n}_{ngeq1}$ is a sequence of real numbers such that $sum_{ngeq1}a_n$ converges. Let $b_n$ be the cardinality of the set ${kin N:a_k>frac{1}{2^n}}$. Show that $limsup_{nrightarrowinfty}frac{1}{2^n}b_n=0$.
My thought is since $sum frac{1}{2^n}$ converges and $sum a_n$ converges, their terms need to be "comparable", but I don't know how to capture this.
real-analysis sequences-and-series
edited Jan 19 at 3:58
Martin Sleziak
45.1k10123277
asked Dec 19 '18 at 11:11
Fluffy SkyeFluffy Skye
34019
$endgroup$
This is the last question on my real analysis final exam. I didn't get to spend much time on it. Hopefully I remember the problem correctly.
${a_n}_{ngeq1}$ is a sequence of real numbers such that $sum_{ngeq1}a_n$ converges. Let $b_n$ be the cardinality of the set ${kin N:a_k>frac{1}{2^n}}$. Show that $limsup_{nrightarrowinfty}frac{1}{2^n}b_n=0$.
My thought is since $sum frac{1}{2^n}$ converges and $sum a_n$ converges, their terms need to be "comparable", but I don't know how to capture this.
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Jan 19 at 3:58
Martin Sleziak
45.1k10123277
asked Dec 19 '18 at 11:11
Fluffy SkyeFluffy Skye
34019
edited Jan 19 at 3:58
Martin Sleziak
45.1k10123277
asked Dec 19 '18 at 11:11
Fluffy SkyeFluffy Skye
34019
edited Jan 19 at 3:58
Martin Sleziak
45.1k10123277
edited Jan 19 at 3:58
Martin Sleziak
45.1k10123277
edited Jan 19 at 3:58
Martin Sleziak
45.1k10123277
45.1k10123277
asked Dec 19 '18 at 11:11
Fluffy SkyeFluffy Skye
34019
asked Dec 19 '18 at 11:11
Fluffy SkyeFluffy Skye
34019
asked Dec 19 '18 at 11:11
Fluffy SkyeFluffy Skye
34019
34019
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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Fix some positive sequence $(c_n)$ with limit $0$, and, for every $n$ and $m$, let $$A_m=sumlimits_{k=m}^infty a_kqquad B_n={kmid a_kgeqslant c_n}$$ Then, $$A_mgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}a_kgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}c_n=c_ncdot#B_ncap[m,infty)$$ hence $$c_ncdot#B_nleqslant c_ncdot m+c_ncdot#B_ncap[m,infty)leqslant c_ncdot m+A_m$$ In the limit $ntoinfty$, one gets, for every fixed $m$, $$limsup_{ntoinfty},c_ncdot#B_nleqslantlimsup_{ntoinfty},(c_ncdot m+A_m)=A_m$$ The series $sum a_k$ converges hence $inflimits_mA_m=0$, thus, $$limsup_{ntoinfty},c_ncdot#B_n=0$$ If $c_n=2^{-n}$ for every $n$, then $#B_n=b_n$ hence we are done.
answered Dec 19 '18 at 12:07
DidDid
249k23228467
$endgroup$
$begingroup$
In the first chain of inequality, should it be $c_ncdot#B_ncap[m,infty)$?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 12:40
$begingroup$
True! Well spotted, thanks.
$endgroup$
– Did
Dec 19 '18 at 12:45
$begingroup$
Emmm, I'm still a bit confused about the last part. You first take n to infinity and then take m to infinity? But then why $#B_m = b_n$ ?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 13:01
$begingroup$
?? $b_n=#B_n$ by definition.
$endgroup$
– Did
Dec 19 '18 at 13:03
$begingroup$
Yeah I know that. But the subscript is different? In the end you have c_n * #B_m not #B_n?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 13:10
|
show 3 more comments
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1 Answer
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1 Answer
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$begingroup$
Fix some positive sequence $(c_n)$ with limit $0$, and, for every $n$ and $m$, let $$A_m=sumlimits_{k=m}^infty a_kqquad B_n={kmid a_kgeqslant c_n}$$ Then, $$A_mgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}a_kgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}c_n=c_ncdot#B_ncap[m,infty)$$ hence $$c_ncdot#B_nleqslant c_ncdot m+c_ncdot#B_ncap[m,infty)leqslant c_ncdot m+A_m$$ In the limit $ntoinfty$, one gets, for every fixed $m$, $$limsup_{ntoinfty},c_ncdot#B_nleqslantlimsup_{ntoinfty},(c_ncdot m+A_m)=A_m$$ The series $sum a_k$ converges hence $inflimits_mA_m=0$, thus, $$limsup_{ntoinfty},c_ncdot#B_n=0$$ If $c_n=2^{-n}$ for every $n$, then $#B_n=b_n$ hence we are done.
answered Dec 19 '18 at 12:07
DidDid
249k23228467
$endgroup$
$begingroup$
In the first chain of inequality, should it be $c_ncdot#B_ncap[m,infty)$?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 12:40
$begingroup$
True! Well spotted, thanks.
$endgroup$
– Did
Dec 19 '18 at 12:45
$begingroup$
Emmm, I'm still a bit confused about the last part. You first take n to infinity and then take m to infinity? But then why $#B_m = b_n$ ?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 13:01
$begingroup$
?? $b_n=#B_n$ by definition.
$endgroup$
– Did
Dec 19 '18 at 13:03
$begingroup$
Yeah I know that. But the subscript is different? In the end you have c_n * #B_m not #B_n?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 13:10
|
show 3 more comments
$begingroup$
Fix some positive sequence $(c_n)$ with limit $0$, and, for every $n$ and $m$, let $$A_m=sumlimits_{k=m}^infty a_kqquad B_n={kmid a_kgeqslant c_n}$$ Then, $$A_mgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}a_kgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}c_n=c_ncdot#B_ncap[m,infty)$$ hence $$c_ncdot#B_nleqslant c_ncdot m+c_ncdot#B_ncap[m,infty)leqslant c_ncdot m+A_m$$ In the limit $ntoinfty$, one gets, for every fixed $m$, $$limsup_{ntoinfty},c_ncdot#B_nleqslantlimsup_{ntoinfty},(c_ncdot m+A_m)=A_m$$ The series $sum a_k$ converges hence $inflimits_mA_m=0$, thus, $$limsup_{ntoinfty},c_ncdot#B_n=0$$ If $c_n=2^{-n}$ for every $n$, then $#B_n=b_n$ hence we are done.
answered Dec 19 '18 at 12:07
DidDid
249k23228467
$endgroup$
$begingroup$
In the first chain of inequality, should it be $c_ncdot#B_ncap[m,infty)$?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 12:40
$begingroup$
True! Well spotted, thanks.
$endgroup$
– Did
Dec 19 '18 at 12:45
$begingroup$
Emmm, I'm still a bit confused about the last part. You first take n to infinity and then take m to infinity? But then why $#B_m = b_n$ ?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 13:01
$begingroup$
?? $b_n=#B_n$ by definition.
$endgroup$
– Did
Dec 19 '18 at 13:03
$begingroup$
Yeah I know that. But the subscript is different? In the end you have c_n * #B_m not #B_n?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 13:10
|
show 3 more comments
$begingroup$
Fix some positive sequence $(c_n)$ with limit $0$, and, for every $n$ and $m$, let $$A_m=sumlimits_{k=m}^infty a_kqquad B_n={kmid a_kgeqslant c_n}$$ Then, $$A_mgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}a_kgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}c_n=c_ncdot#B_ncap[m,infty)$$ hence $$c_ncdot#B_nleqslant c_ncdot m+c_ncdot#B_ncap[m,infty)leqslant c_ncdot m+A_m$$ In the limit $ntoinfty$, one gets, for every fixed $m$, $$limsup_{ntoinfty},c_ncdot#B_nleqslantlimsup_{ntoinfty},(c_ncdot m+A_m)=A_m$$ The series $sum a_k$ converges hence $inflimits_mA_m=0$, thus, $$limsup_{ntoinfty},c_ncdot#B_n=0$$ If $c_n=2^{-n}$ for every $n$, then $#B_n=b_n$ hence we are done.
answered Dec 19 '18 at 12:07
DidDid
249k23228467
$endgroup$
Fix some positive sequence $(c_n)$ with limit $0$, and, for every $n$ and $m$, let $$A_m=sumlimits_{k=m}^infty a_kqquad B_n={kmid a_kgeqslant c_n}$$ Then, $$A_mgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}a_kgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}c_n=c_ncdot#B_ncap[m,infty)$$ hence $$c_ncdot#B_nleqslant c_ncdot m+c_ncdot#B_ncap[m,infty)leqslant c_ncdot m+A_m$$ In the limit $ntoinfty$, one gets, for every fixed $m$, $$limsup_{ntoinfty},c_ncdot#B_nleqslantlimsup_{ntoinfty},(c_ncdot m+A_m)=A_m$$ The series $sum a_k$ converges hence $inflimits_mA_m=0$, thus, $$limsup_{ntoinfty},c_ncdot#B_n=0$$ If $c_n=2^{-n}$ for every $n$, then $#B_n=b_n$ hence we are done.
answered Dec 19 '18 at 12:07
DidDid
249k23228467
edited Dec 19 '18 at 13:59
answered Dec 19 '18 at 12:07
DidDid
249k23228467
answered Dec 19 '18 at 12:07
DidDid
249k23228467
answered Dec 19 '18 at 12:07
DidDid
249k23228467
249k23228467
$begingroup$
In the first chain of inequality, should it be $c_ncdot#B_ncap[m,infty)$?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 12:40
$begingroup$
True! Well spotted, thanks.
$endgroup$
– Did
Dec 19 '18 at 12:45
$begingroup$
Emmm, I'm still a bit confused about the last part. You first take n to infinity and then take m to infinity? But then why $#B_m = b_n$ ?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 13:01
$begingroup$
?? $b_n=#B_n$ by definition.
$endgroup$
– Did
Dec 19 '18 at 13:03
$begingroup$
Yeah I know that. But the subscript is different? In the end you have c_n * #B_m not #B_n?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 13:10
|
show 3 more comments
$begingroup$
In the first chain of inequality, should it be $c_ncdot#B_ncap[m,infty)$?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 12:40
$begingroup$
True! Well spotted, thanks.
$endgroup$
– Did
Dec 19 '18 at 12:45
$begingroup$
Emmm, I'm still a bit confused about the last part. You first take n to infinity and then take m to infinity? But then why $#B_m = b_n$ ?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 13:01
$begingroup$
?? $b_n=#B_n$ by definition.
$endgroup$
– Did
Dec 19 '18 at 13:03
$begingroup$
Yeah I know that. But the subscript is different? In the end you have c_n * #B_m not #B_n?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 13:10
$begingroup$
In the first chain of inequality, should it be $c_ncdot#B_ncap[m,infty)$?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 12:40
$begingroup$
In the first chain of inequality, should it be $c_ncdot#B_ncap[m,infty)$?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 12:40
$begingroup$
True! Well spotted, thanks.
$endgroup$
– Did
Dec 19 '18 at 12:45
$begingroup$
True! Well spotted, thanks.
$endgroup$
– Did
Dec 19 '18 at 12:45
$begingroup$
Emmm, I'm still a bit confused about the last part. You first take n to infinity and then take m to infinity? But then why $#B_m = b_n$ ?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 13:01
$begingroup$
Emmm, I'm still a bit confused about the last part. You first take n to infinity and then take m to infinity? But then why $#B_m = b_n$ ?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 13:01
$begingroup$
?? $b_n=#B_n$ by definition.
$endgroup$
– Did
Dec 19 '18 at 13:03
$begingroup$
?? $b_n=#B_n$ by definition.
$endgroup$
– Did
Dec 19 '18 at 13:03
$begingroup$
Yeah I know that. But the subscript is different? In the end you have c_n * #B_m not #B_n?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 13:10
$begingroup$
Yeah I know that. But the subscript is different? In the end you have c_n * #B_m not #B_n?
$endgroup$
– Fluffy Skye
Dec 19 '18 at 13:10
|
show 3 more comments
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