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Sequence formed by quotient of sum and the product of odd squares is divergent: Why?
Clash Royale CLAN TAG#URR8PPP
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Trust me when I tell you this is not homework. Can you suggest an solid argument to prove
$$ S_n = frac13^25^27^2cdots (2n-1)^2sum_m=3^infty 2^(2n)me^-2^m/2
$$
diverges as $n rightarrow infty$ ?
I have calculated $S_1,S_2,S_3,S_4$ and $S_5$
$$ S_1 = sum_m=3^infty 2^2me^-2^m/2 = 13.63 $$
$$ S_2 = frac13^2sum_m=3^infty 2^4me^-2^m/2 = 1611.5$$
$$ S_3 = frac13^25^2sum_m=3^infty 2^6me^-2^m/2 = 511684$$
$$ S_4 = frac13^25^27^2sum_m=3^infty 2^8me^-2^m/2 = 3.42 times 10^8 $$
$$ S_5 = frac13^25^27^29^2sum_m=3^infty 2^10me^-2^m/2 = frac3.513 times 10^173^25^27^29^2 = 3.934 times 10^11 $$
and I am convinced it diverges, however, I'd like a more solid argument to rest my conjecture $S_n rightarrow infty$ as $n rightarrow infty$.
Thanks in advance. I will award 200pts bounty for a quality answer.
sequences-and-series divergent-series
asked Jul 25 at 19:20
James S. Cook
12.8k22668
 |Â
show 9 more comments
up vote
5
down vote
favorite
Trust me when I tell you this is not homework. Can you suggest an solid argument to prove
$$ S_n = frac13^25^27^2cdots (2n-1)^2sum_m=3^infty 2^(2n)me^-2^m/2
$$
diverges as $n rightarrow infty$ ?
I have calculated $S_1,S_2,S_3,S_4$ and $S_5$
$$ S_1 = sum_m=3^infty 2^2me^-2^m/2 = 13.63 $$
$$ S_2 = frac13^2sum_m=3^infty 2^4me^-2^m/2 = 1611.5$$
$$ S_3 = frac13^25^2sum_m=3^infty 2^6me^-2^m/2 = 511684$$
$$ S_4 = frac13^25^27^2sum_m=3^infty 2^8me^-2^m/2 = 3.42 times 10^8 $$
$$ S_5 = frac13^25^27^29^2sum_m=3^infty 2^10me^-2^m/2 = frac3.513 times 10^173^25^27^29^2 = 3.934 times 10^11 $$
and I am convinced it diverges, however, I'd like a more solid argument to rest my conjecture $S_n rightarrow infty$ as $n rightarrow infty$.
Thanks in advance. I will award 200pts bounty for a quality answer.
sequences-and-series divergent-series
asked Jul 25 at 19:20
James S. Cook
12.8k22668
Can you give your results for $S_1, S_2, S_3, S_4$, and $S_5$? It would help confirm that we have the correct series in mind. My intuition for the series I'm reading coincides with that of @InterstellarProbe, so perhaps I'm not getting the right series.
– Brian Tung
Jul 25 at 19:46
@BrianTung Sure. Just a second I'll edit the results in.
– James S. Cook
Jul 25 at 19:47
@InterstellarProbe do you have a link to that calculation? I can't get wolframalpha to hack it.
– James S. Cook
Jul 25 at 19:59
@JamesS.Cook I figured out why it was giving that result. It was taking the limit first, then doing the summation. So, it was summing zero for every term.
– InterstellarProbe
Jul 25 at 20:01
1
Keeping a single term in the series, chosen such that $2^m/2approx n$, already suffices to show that $S_ngeqslant n^2n+o(n)$ hence $S_ntoinfty$.
– Did
Jul 25 at 20:07
 |Â
show 9 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Trust me when I tell you this is not homework. Can you suggest an solid argument to prove
$$ S_n = frac13^25^27^2cdots (2n-1)^2sum_m=3^infty 2^(2n)me^-2^m/2
$$
diverges as $n rightarrow infty$ ?
I have calculated $S_1,S_2,S_3,S_4$ and $S_5$
$$ S_1 = sum_m=3^infty 2^2me^-2^m/2 = 13.63 $$
$$ S_2 = frac13^2sum_m=3^infty 2^4me^-2^m/2 = 1611.5$$
$$ S_3 = frac13^25^2sum_m=3^infty 2^6me^-2^m/2 = 511684$$
$$ S_4 = frac13^25^27^2sum_m=3^infty 2^8me^-2^m/2 = 3.42 times 10^8 $$
$$ S_5 = frac13^25^27^29^2sum_m=3^infty 2^10me^-2^m/2 = frac3.513 times 10^173^25^27^29^2 = 3.934 times 10^11 $$
and I am convinced it diverges, however, I'd like a more solid argument to rest my conjecture $S_n rightarrow infty$ as $n rightarrow infty$.
Thanks in advance. I will award 200pts bounty for a quality answer.
sequences-and-series divergent-series
asked Jul 25 at 19:20
James S. Cook
12.8k22668
Trust me when I tell you this is not homework. Can you suggest an solid argument to prove
$$ S_n = frac13^25^27^2cdots (2n-1)^2sum_m=3^infty 2^(2n)me^-2^m/2
$$
diverges as $n rightarrow infty$ ?
I have calculated $S_1,S_2,S_3,S_4$ and $S_5$
$$ S_1 = sum_m=3^infty 2^2me^-2^m/2 = 13.63 $$
$$ S_2 = frac13^2sum_m=3^infty 2^4me^-2^m/2 = 1611.5$$
$$ S_3 = frac13^25^2sum_m=3^infty 2^6me^-2^m/2 = 511684$$
$$ S_4 = frac13^25^27^2sum_m=3^infty 2^8me^-2^m/2 = 3.42 times 10^8 $$
$$ S_5 = frac13^25^27^29^2sum_m=3^infty 2^10me^-2^m/2 = frac3.513 times 10^173^25^27^29^2 = 3.934 times 10^11 $$
and I am convinced it diverges, however, I'd like a more solid argument to rest my conjecture $S_n rightarrow infty$ as $n rightarrow infty$.
Thanks in advance. I will award 200pts bounty for a quality answer.
sequences-and-series divergent-series
asked Jul 25 at 19:20
James S. Cook
12.8k22668
edited Jul 25 at 19:48
asked Jul 25 at 19:20
James S. Cook
12.8k22668
asked Jul 25 at 19:20
James S. Cook
12.8k22668
asked Jul 25 at 19:20
James S. Cook
12.8k22668
12.8k22668
Can you give your results for $S_1, S_2, S_3, S_4$, and $S_5$? It would help confirm that we have the correct series in mind. My intuition for the series I'm reading coincides with that of @InterstellarProbe, so perhaps I'm not getting the right series.
– Brian Tung
Jul 25 at 19:46
@BrianTung Sure. Just a second I'll edit the results in.
– James S. Cook
Jul 25 at 19:47
@InterstellarProbe do you have a link to that calculation? I can't get wolframalpha to hack it.
– James S. Cook
Jul 25 at 19:59
@JamesS.Cook I figured out why it was giving that result. It was taking the limit first, then doing the summation. So, it was summing zero for every term.
– InterstellarProbe
Jul 25 at 20:01
1
Keeping a single term in the series, chosen such that $2^m/2approx n$, already suffices to show that $S_ngeqslant n^2n+o(n)$ hence $S_ntoinfty$.
– Did
Jul 25 at 20:07
 |Â
show 9 more comments
Can you give your results for $S_1, S_2, S_3, S_4$, and $S_5$? It would help confirm that we have the correct series in mind. My intuition for the series I'm reading coincides with that of @InterstellarProbe, so perhaps I'm not getting the right series.
– Brian Tung
Jul 25 at 19:46
@BrianTung Sure. Just a second I'll edit the results in.
– James S. Cook
Jul 25 at 19:47
@InterstellarProbe do you have a link to that calculation? I can't get wolframalpha to hack it.
– James S. Cook
Jul 25 at 19:59
@JamesS.Cook I figured out why it was giving that result. It was taking the limit first, then doing the summation. So, it was summing zero for every term.
– InterstellarProbe
Jul 25 at 20:01
1
Keeping a single term in the series, chosen such that $2^m/2approx n$, already suffices to show that $S_ngeqslant n^2n+o(n)$ hence $S_ntoinfty$.
– Did
Jul 25 at 20:07
Can you give your results for $S_1, S_2, S_3, S_4$, and $S_5$? It would help confirm that we have the correct series in mind. My intuition for the series I'm reading coincides with that of @InterstellarProbe, so perhaps I'm not getting the right series.
– Brian Tung
Jul 25 at 19:46
Can you give your results for $S_1, S_2, S_3, S_4$, and $S_5$? It would help confirm that we have the correct series in mind. My intuition for the series I'm reading coincides with that of @InterstellarProbe, so perhaps I'm not getting the right series.
– Brian Tung
Jul 25 at 19:46
@BrianTung Sure. Just a second I'll edit the results in.
– James S. Cook
Jul 25 at 19:47
@BrianTung Sure. Just a second I'll edit the results in.
– James S. Cook
Jul 25 at 19:47
@InterstellarProbe do you have a link to that calculation? I can't get wolframalpha to hack it.
– James S. Cook
Jul 25 at 19:59
@InterstellarProbe do you have a link to that calculation? I can't get wolframalpha to hack it.
– James S. Cook
Jul 25 at 19:59
@JamesS.Cook I figured out why it was giving that result. It was taking the limit first, then doing the summation. So, it was summing zero for every term.
– InterstellarProbe
Jul 25 at 20:01
@JamesS.Cook I figured out why it was giving that result. It was taking the limit first, then doing the summation. So, it was summing zero for every term.
– InterstellarProbe
Jul 25 at 20:01
1
1
Keeping a single term in the series, chosen such that $2^m/2approx n$, already suffices to show that $S_ngeqslant n^2n+o(n)$ hence $S_ntoinfty$.
– Did
Jul 25 at 20:07
Keeping a single term in the series, chosen such that $2^m/2approx n$, already suffices to show that $S_ngeqslant n^2n+o(n)$ hence $S_ntoinfty$.
– Did
Jul 25 at 20:07
 |Â
show 9 more comments
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
This is detailing an observation made by Did in the comments.
Note that, for $ngeq 3$,
$$
S_n = frac1(3cdot 5 cdot 7cdots(2n-1))^2sum_m=3^infty 2^2nm e^-2^m/2
= left(frac2^n n!2n!right)^2sum_m=3^infty 2^2nm e^-2^m/2
$$ hence $$S_ngeq left(frac2^n n!2n!right)^2 max_mgeq 32^2nm e^-2^m/2tag1
$$
This (possibly quite loose) lower bound suggests to choose $m$ as close as possible to the maximizer of the function $f_ncolon[3,infty)tomathbbR$ defined by
$$
f_n(x) = e^2n xln 2-2^x/2,.
$$
Maximizing $f_n$ is equivalent to maximizing $ln f_n$, which is done for $x$ such that $2^x/2 = 4n$. Let
$$2log_2(4n)leq m_nstackrelrm def= lceil 2log_2(4n)rceil < 2log_2(4n)+2$$
so that
$$
4n leq 2^m_n/2< 8ntag2
$$
Then,
$$
S_n geq left(frac2^n n!2n!right)^2 f_n(m_n) geq left(frac2^n n!2n!right)^2 2^4n log_2(4n)e^-8n stackrelrm def= T_ntag3
$$
Now, we rely on Stirling's approximation to bound the first factor of $T_n$:
$$
T_n operatorname*sim_ntoinfty frac12cdotfrace^2n2^2nn^2n cdot 2^4n log_2(4n)e^-8n = 2^4nlog_2 n+O(n) = n^4n+o(n) xrightarrow[ntoinfty] infty tag4
$$
By (3) and (4), we get finally
$$
boxedlim_ntoinftyS_n = infty
$$
edited Jul 26 at 6:39
Did
242k23208442
answered Jul 25 at 22:42
Clement C.
47k33682
Thanks! Your $T_n$ is even more infinite than Did's original comment. Of course, a little infinity was all I needed, most appreciated. It's been a while since I tinkered with Stirling's approximation.
– James S. Cook
Jul 26 at 0:50
Well done (+1).
– Did
Jul 26 at 6:36
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
This is detailing an observation made by Did in the comments.
Note that, for $ngeq 3$,
$$
S_n = frac1(3cdot 5 cdot 7cdots(2n-1))^2sum_m=3^infty 2^2nm e^-2^m/2
= left(frac2^n n!2n!right)^2sum_m=3^infty 2^2nm e^-2^m/2
$$ hence $$S_ngeq left(frac2^n n!2n!right)^2 max_mgeq 32^2nm e^-2^m/2tag1
$$
This (possibly quite loose) lower bound suggests to choose $m$ as close as possible to the maximizer of the function $f_ncolon[3,infty)tomathbbR$ defined by
$$
f_n(x) = e^2n xln 2-2^x/2,.
$$
Maximizing $f_n$ is equivalent to maximizing $ln f_n$, which is done for $x$ such that $2^x/2 = 4n$. Let
$$2log_2(4n)leq m_nstackrelrm def= lceil 2log_2(4n)rceil < 2log_2(4n)+2$$
so that
$$
4n leq 2^m_n/2< 8ntag2
$$
Then,
$$
S_n geq left(frac2^n n!2n!right)^2 f_n(m_n) geq left(frac2^n n!2n!right)^2 2^4n log_2(4n)e^-8n stackrelrm def= T_ntag3
$$
Now, we rely on Stirling's approximation to bound the first factor of $T_n$:
$$
T_n operatorname*sim_ntoinfty frac12cdotfrace^2n2^2nn^2n cdot 2^4n log_2(4n)e^-8n = 2^4nlog_2 n+O(n) = n^4n+o(n) xrightarrow[ntoinfty] infty tag4
$$
By (3) and (4), we get finally
$$
boxedlim_ntoinftyS_n = infty
$$
edited Jul 26 at 6:39
Did
242k23208442
answered Jul 25 at 22:42
Clement C.
47k33682
Thanks! Your $T_n$ is even more infinite than Did's original comment. Of course, a little infinity was all I needed, most appreciated. It's been a while since I tinkered with Stirling's approximation.
– James S. Cook
Jul 26 at 0:50
Well done (+1).
– Did
Jul 26 at 6:36
add a comment |Â
up vote
5
down vote
accepted
This is detailing an observation made by Did in the comments.
Note that, for $ngeq 3$,
$$
S_n = frac1(3cdot 5 cdot 7cdots(2n-1))^2sum_m=3^infty 2^2nm e^-2^m/2
= left(frac2^n n!2n!right)^2sum_m=3^infty 2^2nm e^-2^m/2
$$ hence $$S_ngeq left(frac2^n n!2n!right)^2 max_mgeq 32^2nm e^-2^m/2tag1
$$
This (possibly quite loose) lower bound suggests to choose $m$ as close as possible to the maximizer of the function $f_ncolon[3,infty)tomathbbR$ defined by
$$
f_n(x) = e^2n xln 2-2^x/2,.
$$
Maximizing $f_n$ is equivalent to maximizing $ln f_n$, which is done for $x$ such that $2^x/2 = 4n$. Let
$$2log_2(4n)leq m_nstackrelrm def= lceil 2log_2(4n)rceil < 2log_2(4n)+2$$
so that
$$
4n leq 2^m_n/2< 8ntag2
$$
Then,
$$
S_n geq left(frac2^n n!2n!right)^2 f_n(m_n) geq left(frac2^n n!2n!right)^2 2^4n log_2(4n)e^-8n stackrelrm def= T_ntag3
$$
Now, we rely on Stirling's approximation to bound the first factor of $T_n$:
$$
T_n operatorname*sim_ntoinfty frac12cdotfrace^2n2^2nn^2n cdot 2^4n log_2(4n)e^-8n = 2^4nlog_2 n+O(n) = n^4n+o(n) xrightarrow[ntoinfty] infty tag4
$$
By (3) and (4), we get finally
$$
boxedlim_ntoinftyS_n = infty
$$
edited Jul 26 at 6:39
Did
242k23208442
answered Jul 25 at 22:42
Clement C.
47k33682
Thanks! Your $T_n$ is even more infinite than Did's original comment. Of course, a little infinity was all I needed, most appreciated. It's been a while since I tinkered with Stirling's approximation.
– James S. Cook
Jul 26 at 0:50
Well done (+1).
– Did
Jul 26 at 6:36
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
This is detailing an observation made by Did in the comments.
Note that, for $ngeq 3$,
$$
S_n = frac1(3cdot 5 cdot 7cdots(2n-1))^2sum_m=3^infty 2^2nm e^-2^m/2
= left(frac2^n n!2n!right)^2sum_m=3^infty 2^2nm e^-2^m/2
$$ hence $$S_ngeq left(frac2^n n!2n!right)^2 max_mgeq 32^2nm e^-2^m/2tag1
$$
This (possibly quite loose) lower bound suggests to choose $m$ as close as possible to the maximizer of the function $f_ncolon[3,infty)tomathbbR$ defined by
$$
f_n(x) = e^2n xln 2-2^x/2,.
$$
Maximizing $f_n$ is equivalent to maximizing $ln f_n$, which is done for $x$ such that $2^x/2 = 4n$. Let
$$2log_2(4n)leq m_nstackrelrm def= lceil 2log_2(4n)rceil < 2log_2(4n)+2$$
so that
$$
4n leq 2^m_n/2< 8ntag2
$$
Then,
$$
S_n geq left(frac2^n n!2n!right)^2 f_n(m_n) geq left(frac2^n n!2n!right)^2 2^4n log_2(4n)e^-8n stackrelrm def= T_ntag3
$$
Now, we rely on Stirling's approximation to bound the first factor of $T_n$:
$$
T_n operatorname*sim_ntoinfty frac12cdotfrace^2n2^2nn^2n cdot 2^4n log_2(4n)e^-8n = 2^4nlog_2 n+O(n) = n^4n+o(n) xrightarrow[ntoinfty] infty tag4
$$
By (3) and (4), we get finally
$$
boxedlim_ntoinftyS_n = infty
$$
edited Jul 26 at 6:39
Did
242k23208442
answered Jul 25 at 22:42
Clement C.
47k33682
This is detailing an observation made by Did in the comments.
Note that, for $ngeq 3$,
$$
S_n = frac1(3cdot 5 cdot 7cdots(2n-1))^2sum_m=3^infty 2^2nm e^-2^m/2
= left(frac2^n n!2n!right)^2sum_m=3^infty 2^2nm e^-2^m/2
$$ hence $$S_ngeq left(frac2^n n!2n!right)^2 max_mgeq 32^2nm e^-2^m/2tag1
$$
This (possibly quite loose) lower bound suggests to choose $m$ as close as possible to the maximizer of the function $f_ncolon[3,infty)tomathbbR$ defined by
$$
f_n(x) = e^2n xln 2-2^x/2,.
$$
Maximizing $f_n$ is equivalent to maximizing $ln f_n$, which is done for $x$ such that $2^x/2 = 4n$. Let
$$2log_2(4n)leq m_nstackrelrm def= lceil 2log_2(4n)rceil < 2log_2(4n)+2$$
so that
$$
4n leq 2^m_n/2< 8ntag2
$$
Then,
$$
S_n geq left(frac2^n n!2n!right)^2 f_n(m_n) geq left(frac2^n n!2n!right)^2 2^4n log_2(4n)e^-8n stackrelrm def= T_ntag3
$$
Now, we rely on Stirling's approximation to bound the first factor of $T_n$:
$$
T_n operatorname*sim_ntoinfty frac12cdotfrace^2n2^2nn^2n cdot 2^4n log_2(4n)e^-8n = 2^4nlog_2 n+O(n) = n^4n+o(n) xrightarrow[ntoinfty] infty tag4
$$
By (3) and (4), we get finally
$$
boxedlim_ntoinftyS_n = infty
$$
edited Jul 26 at 6:39
Did
242k23208442
answered Jul 25 at 22:42
Clement C.
47k33682
edited Jul 26 at 6:39
Did
242k23208442
edited Jul 26 at 6:39
Did
242k23208442
edited Jul 26 at 6:39
Did
242k23208442
242k23208442
answered Jul 25 at 22:42
Clement C.
47k33682
answered Jul 25 at 22:42
Clement C.
47k33682
answered Jul 25 at 22:42
Clement C.
47k33682
47k33682
Thanks! Your $T_n$ is even more infinite than Did's original comment. Of course, a little infinity was all I needed, most appreciated. It's been a while since I tinkered with Stirling's approximation.
– James S. Cook
Jul 26 at 0:50
Well done (+1).
– Did
Jul 26 at 6:36
add a comment |Â
Thanks! Your $T_n$ is even more infinite than Did's original comment. Of course, a little infinity was all I needed, most appreciated. It's been a while since I tinkered with Stirling's approximation.
– James S. Cook
Jul 26 at 0:50
Well done (+1).
– Did
Jul 26 at 6:36
Thanks! Your $T_n$ is even more infinite than Did's original comment. Of course, a little infinity was all I needed, most appreciated. It's been a while since I tinkered with Stirling's approximation.
– James S. Cook
Jul 26 at 0:50
Thanks! Your $T_n$ is even more infinite than Did's original comment. Of course, a little infinity was all I needed, most appreciated. It's been a while since I tinkered with Stirling's approximation.
– James S. Cook
Jul 26 at 0:50
Well done (+1).
– Did
Jul 26 at 6:36
Well done (+1).
– Did
Jul 26 at 6:36
add a comment |Â
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Can you give your results for $S_1, S_2, S_3, S_4$, and $S_5$? It would help confirm that we have the correct series in mind. My intuition for the series I'm reading coincides with that of @InterstellarProbe, so perhaps I'm not getting the right series.
– Brian Tung
Jul 25 at 19:46
@BrianTung Sure. Just a second I'll edit the results in.
– James S. Cook
Jul 25 at 19:47
@InterstellarProbe do you have a link to that calculation? I can't get wolframalpha to hack it.
– James S. Cook
Jul 25 at 19:59
@JamesS.Cook I figured out why it was giving that result. It was taking the limit first, then doing the summation. So, it was summing zero for every term.
– InterstellarProbe
Jul 25 at 20:01
1
Keeping a single term in the series, chosen such that $2^m/2approx n$, already suffices to show that $S_ngeqslant n^2n+o(n)$ hence $S_ntoinfty$.
– Did
Jul 25 at 20:07