Mixing
If a positive increasing sequence tends to infinity, then $sum_{n=2}^infty frac{a_n-a_{n-1}}{a_n}=infty$
$begingroup$
While preparing my tutoring group on (currently) measure theory and Lebesgue integration on $mathbb R^n$, I by chance came across the following result (at least I hope this is a result and my proof is valid) which I don't recall seeing up until today.
Let $(a_n)_{ninmathbb N}$ be a monotonically increasing sequence of positive numbers such that $lim_{ntoinfty}a_n=infty$. Then
$$sum_{n=2}^infty frac{a_n-a_{n-1}}{a_n}=infty,.$$
The proof I found for this relies on dominated convergence and goes as follows:
Proof. Given the sequence $(a_n)_{ninmathbb N}$ in question, define a sequence of functions $(f_n)_{ninmathbb N}$ via
$$
f_n:[0,infty)to [0,infty)qquad xmapsto frac1{a_n}mathbb 1_{[0,a_n]}(x)
$$
with $mathbb 1_{A}$ being the usual indicator function for arbitrary $Asubseteqmathbb R$. Obviously, $int_0^infty f_n(x),dx=1$ for all $ninmathbb N$ and $(f_n)_{ninmathbb N}$ converges pointwise to zero as $lim_{ntoinfty}a_n=infty$ by assumption. This however implies
$$
lim_{ntoinfty}int_0^infty f_n(x),dx=1neq 0=int_0^infty0,dx=int_0^inftylim_{ntoinfty} f_n(x),dx
$$
so by (the converse of) Lebesgue's dominated convergence theorem, no dominating integrable function $g$ for $(f_n)_{ninmathbb N}$ can exist. This in turn means that every dominating function $g$ we find has to be non-integrable. The obvious choice here is (written in a sloppy way but it should be clear how $g$ operates)
$$
g:[0,infty)to[0,infty)qquad xmapsto begin{cases} frac1{a_1}&text{ if }xin[0,a_1]\frac1{a_2}&text{ if }xin(a_1,a_2]\cdots end{cases},.
$$
Note that $g$ is well-defined as $(a_n)_n$ is assumed to be monotonically increasing and unconditionally convergent and, evidently, $|f_n|leq g$ for all $ninmathbb N$. Finally,
$$
infty=int_0^infty g(x),dx=operatorname{vol}([0,a_1])cdotfrac1{a_1}+operatorname{vol}((a_1,a_2])cdotfrac1{a_2}+ldots=1+sum_{n=2}^infty frac{a_n-a_{n-1}}{a_n}
$$
which concludes the proof.$quadsquare$
Of course the above result is trivial if $lim_{ntoinfty}frac{a_{n-1}}{a_n}neq 1$, but assuming $(a_n-a_{n-1})/a_nto 0$ as $ntoinfty$, I did not see an easy way to obtain this in an elementary way - admittedly, I didn't think about an alternate proof for too long. Given a quick glance I also did not find a similar result on this site just yet. Hence my question is:
Is there a simpler way to show this result (assuming it holds and I did not make a mistake)? If so, was this maybe already discussed somewhere on math.SE?
Thanks in advance for any answer or comment!
As a final remark (or rather small example), a direct application of this to the sequence $a_n:=sum_{k=1}^n frac1k$ immediatly yields the divergence of the series
$$
sum_{n=2}^infty frac1{n(sum_{k=1}^n frac1k)}=infty,.
$$
Given the divergence of $sum_{n=2}^infty frac1{nlog(n)}$ this is of course not surprising but nontheless this (at least in my opinion) is interesting to play around with.
real-analysis calculus sequences-and-series proof-verification lebesgue-integral
asked Jan 7 at 13:21
Frederik vom EndeFrederik vom Ende
6621321
$endgroup$
add a comment |
$begingroup$
While preparing my tutoring group on (currently) measure theory and Lebesgue integration on $mathbb R^n$, I by chance came across the following result (at least I hope this is a result and my proof is valid) which I don't recall seeing up until today.
Let $(a_n)_{ninmathbb N}$ be a monotonically increasing sequence of positive numbers such that $lim_{ntoinfty}a_n=infty$. Then
$$sum_{n=2}^infty frac{a_n-a_{n-1}}{a_n}=infty,.$$
The proof I found for this relies on dominated convergence and goes as follows:
Proof. Given the sequence $(a_n)_{ninmathbb N}$ in question, define a sequence of functions $(f_n)_{ninmathbb N}$ via
$$
f_n:[0,infty)to [0,infty)qquad xmapsto frac1{a_n}mathbb 1_{[0,a_n]}(x)
$$
with $mathbb 1_{A}$ being the usual indicator function for arbitrary $Asubseteqmathbb R$. Obviously, $int_0^infty f_n(x),dx=1$ for all $ninmathbb N$ and $(f_n)_{ninmathbb N}$ converges pointwise to zero as $lim_{ntoinfty}a_n=infty$ by assumption. This however implies
$$
lim_{ntoinfty}int_0^infty f_n(x),dx=1neq 0=int_0^infty0,dx=int_0^inftylim_{ntoinfty} f_n(x),dx
$$
so by (the converse of) Lebesgue's dominated convergence theorem, no dominating integrable function $g$ for $(f_n)_{ninmathbb N}$ can exist. This in turn means that every dominating function $g$ we find has to be non-integrable. The obvious choice here is (written in a sloppy way but it should be clear how $g$ operates)
$$
g:[0,infty)to[0,infty)qquad xmapsto begin{cases} frac1{a_1}&text{ if }xin[0,a_1]\frac1{a_2}&text{ if }xin(a_1,a_2]\cdots end{cases},.
$$
Note that $g$ is well-defined as $(a_n)_n$ is assumed to be monotonically increasing and unconditionally convergent and, evidently, $|f_n|leq g$ for all $ninmathbb N$. Finally,
$$
infty=int_0^infty g(x),dx=operatorname{vol}([0,a_1])cdotfrac1{a_1}+operatorname{vol}((a_1,a_2])cdotfrac1{a_2}+ldots=1+sum_{n=2}^infty frac{a_n-a_{n-1}}{a_n}
$$
which concludes the proof.$quadsquare$
Of course the above result is trivial if $lim_{ntoinfty}frac{a_{n-1}}{a_n}neq 1$, but assuming $(a_n-a_{n-1})/a_nto 0$ as $ntoinfty$, I did not see an easy way to obtain this in an elementary way - admittedly, I didn't think about an alternate proof for too long. Given a quick glance I also did not find a similar result on this site just yet. Hence my question is:
Is there a simpler way to show this result (assuming it holds and I did not make a mistake)? If so, was this maybe already discussed somewhere on math.SE?
Thanks in advance for any answer or comment!
As a final remark (or rather small example), a direct application of this to the sequence $a_n:=sum_{k=1}^n frac1k$ immediatly yields the divergence of the series
$$
sum_{n=2}^infty frac1{n(sum_{k=1}^n frac1k)}=infty,.
$$
Given the divergence of $sum_{n=2}^infty frac1{nlog(n)}$ this is of course not surprising but nontheless this (at least in my opinion) is interesting to play around with.
real-analysis calculus sequences-and-series proof-verification lebesgue-integral
asked Jan 7 at 13:21
Frederik vom EndeFrederik vom Ende
6621321
$endgroup$
1
$begingroup$
Simpler (?) proofs here math.stackexchange.com/q/746257/42969 and here math.stackexchange.com/q/388898/42969.
$endgroup$
– Martin R
Jan 7 at 13:31
$begingroup$
@Henry Changed the title to get rid of the ambiguous wording, thank you!
$endgroup$
– Frederik vom Ende
Jan 7 at 13:51
$begingroup$
@MartinR I'm not sure if the proof in question is directly applicable as the denominator differs from the one used here, i.e. $(a_{n+1}-a_n)/a_n$ vs. $(a_{n+1}-a_n)/a_{n+1}$ (with the latter being smaller so the above result should even be slightly stronger than the linked one?)
$endgroup$
– Frederik vom Ende
Jan 7 at 14:17
$begingroup$
Using lowercase letters for your sequence, and uppercase letters for the sequences in math.stackexchange.com/q/388898/42969, we have $a_n = S_n$, $a_n - a_{n-1} = S_n - S_{n-1} = A_n$, and therefore the identical series$ sum frac{a_n-a_{n-1}}{a_n} =sum frac{A_n}{S_n}$. Or did I make some error?
$endgroup$
– Martin R
Jan 7 at 14:22
$begingroup$
Yeah you are of course correct; my annotation rather was related to the post I linked in my comment (which seems to not be equivalent to, but implied by the above result). Anyways, thank you for your time and the useful links!
$endgroup$
– Frederik vom Ende
Jan 7 at 14:33
add a comment |
$begingroup$
While preparing my tutoring group on (currently) measure theory and Lebesgue integration on $mathbb R^n$, I by chance came across the following result (at least I hope this is a result and my proof is valid) which I don't recall seeing up until today.
Let $(a_n)_{ninmathbb N}$ be a monotonically increasing sequence of positive numbers such that $lim_{ntoinfty}a_n=infty$. Then
$$sum_{n=2}^infty frac{a_n-a_{n-1}}{a_n}=infty,.$$
The proof I found for this relies on dominated convergence and goes as follows:
Proof. Given the sequence $(a_n)_{ninmathbb N}$ in question, define a sequence of functions $(f_n)_{ninmathbb N}$ via
$$
f_n:[0,infty)to [0,infty)qquad xmapsto frac1{a_n}mathbb 1_{[0,a_n]}(x)
$$
with $mathbb 1_{A}$ being the usual indicator function for arbitrary $Asubseteqmathbb R$. Obviously, $int_0^infty f_n(x),dx=1$ for all $ninmathbb N$ and $(f_n)_{ninmathbb N}$ converges pointwise to zero as $lim_{ntoinfty}a_n=infty$ by assumption. This however implies
$$
lim_{ntoinfty}int_0^infty f_n(x),dx=1neq 0=int_0^infty0,dx=int_0^inftylim_{ntoinfty} f_n(x),dx
$$
so by (the converse of) Lebesgue's dominated convergence theorem, no dominating integrable function $g$ for $(f_n)_{ninmathbb N}$ can exist. This in turn means that every dominating function $g$ we find has to be non-integrable. The obvious choice here is (written in a sloppy way but it should be clear how $g$ operates)
$$
g:[0,infty)to[0,infty)qquad xmapsto begin{cases} frac1{a_1}&text{ if }xin[0,a_1]\frac1{a_2}&text{ if }xin(a_1,a_2]\cdots end{cases},.
$$
Note that $g$ is well-defined as $(a_n)_n$ is assumed to be monotonically increasing and unconditionally convergent and, evidently, $|f_n|leq g$ for all $ninmathbb N$. Finally,
$$
infty=int_0^infty g(x),dx=operatorname{vol}([0,a_1])cdotfrac1{a_1}+operatorname{vol}((a_1,a_2])cdotfrac1{a_2}+ldots=1+sum_{n=2}^infty frac{a_n-a_{n-1}}{a_n}
$$
which concludes the proof.$quadsquare$
Of course the above result is trivial if $lim_{ntoinfty}frac{a_{n-1}}{a_n}neq 1$, but assuming $(a_n-a_{n-1})/a_nto 0$ as $ntoinfty$, I did not see an easy way to obtain this in an elementary way - admittedly, I didn't think about an alternate proof for too long. Given a quick glance I also did not find a similar result on this site just yet. Hence my question is:
Is there a simpler way to show this result (assuming it holds and I did not make a mistake)? If so, was this maybe already discussed somewhere on math.SE?
Thanks in advance for any answer or comment!
As a final remark (or rather small example), a direct application of this to the sequence $a_n:=sum_{k=1}^n frac1k$ immediatly yields the divergence of the series
$$
sum_{n=2}^infty frac1{n(sum_{k=1}^n frac1k)}=infty,.
$$
Given the divergence of $sum_{n=2}^infty frac1{nlog(n)}$ this is of course not surprising but nontheless this (at least in my opinion) is interesting to play around with.
real-analysis calculus sequences-and-series proof-verification lebesgue-integral
asked Jan 7 at 13:21
Frederik vom EndeFrederik vom Ende
6621321
$endgroup$
While preparing my tutoring group on (currently) measure theory and Lebesgue integration on $mathbb R^n$, I by chance came across the following result (at least I hope this is a result and my proof is valid) which I don't recall seeing up until today.
Let $(a_n)_{ninmathbb N}$ be a monotonically increasing sequence of positive numbers such that $lim_{ntoinfty}a_n=infty$. Then
$$sum_{n=2}^infty frac{a_n-a_{n-1}}{a_n}=infty,.$$
The proof I found for this relies on dominated convergence and goes as follows:
Proof. Given the sequence $(a_n)_{ninmathbb N}$ in question, define a sequence of functions $(f_n)_{ninmathbb N}$ via
$$
f_n:[0,infty)to [0,infty)qquad xmapsto frac1{a_n}mathbb 1_{[0,a_n]}(x)
$$
with $mathbb 1_{A}$ being the usual indicator function for arbitrary $Asubseteqmathbb R$. Obviously, $int_0^infty f_n(x),dx=1$ for all $ninmathbb N$ and $(f_n)_{ninmathbb N}$ converges pointwise to zero as $lim_{ntoinfty}a_n=infty$ by assumption. This however implies
$$
lim_{ntoinfty}int_0^infty f_n(x),dx=1neq 0=int_0^infty0,dx=int_0^inftylim_{ntoinfty} f_n(x),dx
$$
so by (the converse of) Lebesgue's dominated convergence theorem, no dominating integrable function $g$ for $(f_n)_{ninmathbb N}$ can exist. This in turn means that every dominating function $g$ we find has to be non-integrable. The obvious choice here is (written in a sloppy way but it should be clear how $g$ operates)
$$
g:[0,infty)to[0,infty)qquad xmapsto begin{cases} frac1{a_1}&text{ if }xin[0,a_1]\frac1{a_2}&text{ if }xin(a_1,a_2]\cdots end{cases},.
$$
Note that $g$ is well-defined as $(a_n)_n$ is assumed to be monotonically increasing and unconditionally convergent and, evidently, $|f_n|leq g$ for all $ninmathbb N$. Finally,
$$
infty=int_0^infty g(x),dx=operatorname{vol}([0,a_1])cdotfrac1{a_1}+operatorname{vol}((a_1,a_2])cdotfrac1{a_2}+ldots=1+sum_{n=2}^infty frac{a_n-a_{n-1}}{a_n}
$$
which concludes the proof.$quadsquare$
Of course the above result is trivial if $lim_{ntoinfty}frac{a_{n-1}}{a_n}neq 1$, but assuming $(a_n-a_{n-1})/a_nto 0$ as $ntoinfty$, I did not see an easy way to obtain this in an elementary way - admittedly, I didn't think about an alternate proof for too long. Given a quick glance I also did not find a similar result on this site just yet. Hence my question is:
Is there a simpler way to show this result (assuming it holds and I did not make a mistake)? If so, was this maybe already discussed somewhere on math.SE?
Thanks in advance for any answer or comment!
As a final remark (or rather small example), a direct application of this to the sequence $a_n:=sum_{k=1}^n frac1k$ immediatly yields the divergence of the series
$$
sum_{n=2}^infty frac1{n(sum_{k=1}^n frac1k)}=infty,.
$$
Given the divergence of $sum_{n=2}^infty frac1{nlog(n)}$ this is of course not surprising but nontheless this (at least in my opinion) is interesting to play around with.
real-analysis calculus sequences-and-series proof-verification lebesgue-integral
real-analysis calculus sequences-and-series proof-verification lebesgue-integral
asked Jan 7 at 13:21
Frederik vom EndeFrederik vom Ende
6621321
asked Jan 7 at 13:21
Frederik vom EndeFrederik vom Ende
6621321
edited Jan 7 at 13:50
Frederik vom Ende
asked Jan 7 at 13:21
Frederik vom EndeFrederik vom Ende
6621321
asked Jan 7 at 13:21
Frederik vom EndeFrederik vom Ende
6621321
asked Jan 7 at 13:21
Frederik vom EndeFrederik vom Ende
6621321
6621321
1
$begingroup$
Simpler (?) proofs here math.stackexchange.com/q/746257/42969 and here math.stackexchange.com/q/388898/42969.
$endgroup$
– Martin R
Jan 7 at 13:31
$begingroup$
@Henry Changed the title to get rid of the ambiguous wording, thank you!
$endgroup$
– Frederik vom Ende
Jan 7 at 13:51
$begingroup$
@MartinR I'm not sure if the proof in question is directly applicable as the denominator differs from the one used here, i.e. $(a_{n+1}-a_n)/a_n$ vs. $(a_{n+1}-a_n)/a_{n+1}$ (with the latter being smaller so the above result should even be slightly stronger than the linked one?)
$endgroup$
– Frederik vom Ende
Jan 7 at 14:17
$begingroup$
Using lowercase letters for your sequence, and uppercase letters for the sequences in math.stackexchange.com/q/388898/42969, we have $a_n = S_n$, $a_n - a_{n-1} = S_n - S_{n-1} = A_n$, and therefore the identical series$ sum frac{a_n-a_{n-1}}{a_n} =sum frac{A_n}{S_n}$. Or did I make some error?
$endgroup$
– Martin R
Jan 7 at 14:22
$begingroup$
Yeah you are of course correct; my annotation rather was related to the post I linked in my comment (which seems to not be equivalent to, but implied by the above result). Anyways, thank you for your time and the useful links!
$endgroup$
– Frederik vom Ende
Jan 7 at 14:33
add a comment |
1
$begingroup$
Simpler (?) proofs here math.stackexchange.com/q/746257/42969 and here math.stackexchange.com/q/388898/42969.
$endgroup$
– Martin R
Jan 7 at 13:31
$begingroup$
@Henry Changed the title to get rid of the ambiguous wording, thank you!
$endgroup$
– Frederik vom Ende
Jan 7 at 13:51
$begingroup$
@MartinR I'm not sure if the proof in question is directly applicable as the denominator differs from the one used here, i.e. $(a_{n+1}-a_n)/a_n$ vs. $(a_{n+1}-a_n)/a_{n+1}$ (with the latter being smaller so the above result should even be slightly stronger than the linked one?)
$endgroup$
– Frederik vom Ende
Jan 7 at 14:17
$begingroup$
Using lowercase letters for your sequence, and uppercase letters for the sequences in math.stackexchange.com/q/388898/42969, we have $a_n = S_n$, $a_n - a_{n-1} = S_n - S_{n-1} = A_n$, and therefore the identical series$ sum frac{a_n-a_{n-1}}{a_n} =sum frac{A_n}{S_n}$. Or did I make some error?
$endgroup$
– Martin R
Jan 7 at 14:22
$begingroup$
Yeah you are of course correct; my annotation rather was related to the post I linked in my comment (which seems to not be equivalent to, but implied by the above result). Anyways, thank you for your time and the useful links!
$endgroup$
– Frederik vom Ende
Jan 7 at 14:33
1
1
$begingroup$
Simpler (?) proofs here math.stackexchange.com/q/746257/42969 and here math.stackexchange.com/q/388898/42969.
$endgroup$
– Martin R
Jan 7 at 13:31
$begingroup$
Simpler (?) proofs here math.stackexchange.com/q/746257/42969 and here math.stackexchange.com/q/388898/42969.
$endgroup$
– Martin R
Jan 7 at 13:31
$begingroup$
@Henry Changed the title to get rid of the ambiguous wording, thank you!
$endgroup$
– Frederik vom Ende
Jan 7 at 13:51
$begingroup$
@Henry Changed the title to get rid of the ambiguous wording, thank you!
$endgroup$
– Frederik vom Ende
Jan 7 at 13:51
$begingroup$
@MartinR I'm not sure if the proof in question is directly applicable as the denominator differs from the one used here, i.e. $(a_{n+1}-a_n)/a_n$ vs. $(a_{n+1}-a_n)/a_{n+1}$ (with the latter being smaller so the above result should even be slightly stronger than the linked one?)
$endgroup$
– Frederik vom Ende
Jan 7 at 14:17
$begingroup$
@MartinR I'm not sure if the proof in question is directly applicable as the denominator differs from the one used here, i.e. $(a_{n+1}-a_n)/a_n$ vs. $(a_{n+1}-a_n)/a_{n+1}$ (with the latter being smaller so the above result should even be slightly stronger than the linked one?)
$endgroup$
– Frederik vom Ende
Jan 7 at 14:17
$begingroup$
Using lowercase letters for your sequence, and uppercase letters for the sequences in math.stackexchange.com/q/388898/42969, we have $a_n = S_n$, $a_n - a_{n-1} = S_n - S_{n-1} = A_n$, and therefore the identical series$ sum frac{a_n-a_{n-1}}{a_n} =sum frac{A_n}{S_n}$. Or did I make some error?
$endgroup$
– Martin R
Jan 7 at 14:22
$begingroup$
Using lowercase letters for your sequence, and uppercase letters for the sequences in math.stackexchange.com/q/388898/42969, we have $a_n = S_n$, $a_n - a_{n-1} = S_n - S_{n-1} = A_n$, and therefore the identical series$ sum frac{a_n-a_{n-1}}{a_n} =sum frac{A_n}{S_n}$. Or did I make some error?
$endgroup$
– Martin R
Jan 7 at 14:22
$begingroup$
Yeah you are of course correct; my annotation rather was related to the post I linked in my comment (which seems to not be equivalent to, but implied by the above result). Anyways, thank you for your time and the useful links!
$endgroup$
– Frederik vom Ende
Jan 7 at 14:33
$begingroup$
Yeah you are of course correct; my annotation rather was related to the post I linked in my comment (which seems to not be equivalent to, but implied by the above result). Anyways, thank you for your time and the useful links!
$endgroup$
– Frederik vom Ende
Jan 7 at 14:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Heuristics. Consider its continuum analogue: If $y(t) > 0$ and $y(t) to infty$ as $t to infty$, then
$$ int_{0}^{R} frac{y'(t)}{y(t)} , mathrm{d}t = log y(R) - log y(0) xrightarrow[Rtoinfty]{} infty.$$
We can adapt this intuition to our case. Write $b_n = (a_n - a_{n-1})/a_n$. Then it suffices to assume that $b_n to 0$, for otherwise the conclusion is trivial. Now since
$$ lim_{ntoinfty} frac{b_n}{log a_n - log a_{n-1}} = lim_{ntoinfty} frac{b_n}{-log(1 - b_n)} = 1, $$
the conclusion follows from
(Limit Comparison Test) If $A_n, B_n > 0$ and $lim A_n/B_n $ converges in $(0, infty)$, then $sum_n A_n$ converges if and only if $sum_n B_n$ converges.
$sum_{n=1}^{N} (log a_n - log a_{n-1}) = log a_N - log a_0 to infty$ as $Ntoinfty$, provided $lim a_n = infty$.
answered Jan 7 at 14:15
Sangchul LeeSangchul Lee
92.7k12167269
$endgroup$
$begingroup$
A truly splendid argument, thank you!
$endgroup$
– Frederik vom Ende
Jan 7 at 14:27
1
$begingroup$
@FrederikvomEnde, Glad it helps! I realized that my argument assumes $b_n > 0$ for all $n$, but this causes no harm as we can remove repetitions from $(a_n)$, i.e., we have $$sum_{n=2}^{infty} left(1 - frac{a_{n-1}}{a_n} right) = sum_{k=1}^{infty} left(1 - frac{a_{n_{k-1}}}{a_{n_k}} right) $$ where $n = n_k$ is the $k$-th smallest index at which $a_{n-1} < a_n$ holds (with $n_0 := 1$).
$endgroup$
– Sangchul Lee
Jan 7 at 14:33
add a comment |
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1 Answer
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$begingroup$
Heuristics. Consider its continuum analogue: If $y(t) > 0$ and $y(t) to infty$ as $t to infty$, then
$$ int_{0}^{R} frac{y'(t)}{y(t)} , mathrm{d}t = log y(R) - log y(0) xrightarrow[Rtoinfty]{} infty.$$
We can adapt this intuition to our case. Write $b_n = (a_n - a_{n-1})/a_n$. Then it suffices to assume that $b_n to 0$, for otherwise the conclusion is trivial. Now since
$$ lim_{ntoinfty} frac{b_n}{log a_n - log a_{n-1}} = lim_{ntoinfty} frac{b_n}{-log(1 - b_n)} = 1, $$
the conclusion follows from
(Limit Comparison Test) If $A_n, B_n > 0$ and $lim A_n/B_n $ converges in $(0, infty)$, then $sum_n A_n$ converges if and only if $sum_n B_n$ converges.
$sum_{n=1}^{N} (log a_n - log a_{n-1}) = log a_N - log a_0 to infty$ as $Ntoinfty$, provided $lim a_n = infty$.
answered Jan 7 at 14:15
Sangchul LeeSangchul Lee
92.7k12167269
$endgroup$
$begingroup$
A truly splendid argument, thank you!
$endgroup$
– Frederik vom Ende
Jan 7 at 14:27
1
$begingroup$
@FrederikvomEnde, Glad it helps! I realized that my argument assumes $b_n > 0$ for all $n$, but this causes no harm as we can remove repetitions from $(a_n)$, i.e., we have $$sum_{n=2}^{infty} left(1 - frac{a_{n-1}}{a_n} right) = sum_{k=1}^{infty} left(1 - frac{a_{n_{k-1}}}{a_{n_k}} right) $$ where $n = n_k$ is the $k$-th smallest index at which $a_{n-1} < a_n$ holds (with $n_0 := 1$).
$endgroup$
– Sangchul Lee
Jan 7 at 14:33
add a comment |
$begingroup$
Heuristics. Consider its continuum analogue: If $y(t) > 0$ and $y(t) to infty$ as $t to infty$, then
$$ int_{0}^{R} frac{y'(t)}{y(t)} , mathrm{d}t = log y(R) - log y(0) xrightarrow[Rtoinfty]{} infty.$$
We can adapt this intuition to our case. Write $b_n = (a_n - a_{n-1})/a_n$. Then it suffices to assume that $b_n to 0$, for otherwise the conclusion is trivial. Now since
$$ lim_{ntoinfty} frac{b_n}{log a_n - log a_{n-1}} = lim_{ntoinfty} frac{b_n}{-log(1 - b_n)} = 1, $$
the conclusion follows from
(Limit Comparison Test) If $A_n, B_n > 0$ and $lim A_n/B_n $ converges in $(0, infty)$, then $sum_n A_n$ converges if and only if $sum_n B_n$ converges.
$sum_{n=1}^{N} (log a_n - log a_{n-1}) = log a_N - log a_0 to infty$ as $Ntoinfty$, provided $lim a_n = infty$.
answered Jan 7 at 14:15
Sangchul LeeSangchul Lee
92.7k12167269
$endgroup$
$begingroup$
A truly splendid argument, thank you!
$endgroup$
– Frederik vom Ende
Jan 7 at 14:27
1
$begingroup$
@FrederikvomEnde, Glad it helps! I realized that my argument assumes $b_n > 0$ for all $n$, but this causes no harm as we can remove repetitions from $(a_n)$, i.e., we have $$sum_{n=2}^{infty} left(1 - frac{a_{n-1}}{a_n} right) = sum_{k=1}^{infty} left(1 - frac{a_{n_{k-1}}}{a_{n_k}} right) $$ where $n = n_k$ is the $k$-th smallest index at which $a_{n-1} < a_n$ holds (with $n_0 := 1$).
$endgroup$
– Sangchul Lee
Jan 7 at 14:33
add a comment |
$begingroup$
Heuristics. Consider its continuum analogue: If $y(t) > 0$ and $y(t) to infty$ as $t to infty$, then
$$ int_{0}^{R} frac{y'(t)}{y(t)} , mathrm{d}t = log y(R) - log y(0) xrightarrow[Rtoinfty]{} infty.$$
We can adapt this intuition to our case. Write $b_n = (a_n - a_{n-1})/a_n$. Then it suffices to assume that $b_n to 0$, for otherwise the conclusion is trivial. Now since
$$ lim_{ntoinfty} frac{b_n}{log a_n - log a_{n-1}} = lim_{ntoinfty} frac{b_n}{-log(1 - b_n)} = 1, $$
the conclusion follows from
(Limit Comparison Test) If $A_n, B_n > 0$ and $lim A_n/B_n $ converges in $(0, infty)$, then $sum_n A_n$ converges if and only if $sum_n B_n$ converges.
$sum_{n=1}^{N} (log a_n - log a_{n-1}) = log a_N - log a_0 to infty$ as $Ntoinfty$, provided $lim a_n = infty$.
answered Jan 7 at 14:15
Sangchul LeeSangchul Lee
92.7k12167269
$endgroup$
Heuristics. Consider its continuum analogue: If $y(t) > 0$ and $y(t) to infty$ as $t to infty$, then
$$ int_{0}^{R} frac{y'(t)}{y(t)} , mathrm{d}t = log y(R) - log y(0) xrightarrow[Rtoinfty]{} infty.$$
We can adapt this intuition to our case. Write $b_n = (a_n - a_{n-1})/a_n$. Then it suffices to assume that $b_n to 0$, for otherwise the conclusion is trivial. Now since
$$ lim_{ntoinfty} frac{b_n}{log a_n - log a_{n-1}} = lim_{ntoinfty} frac{b_n}{-log(1 - b_n)} = 1, $$
the conclusion follows from
(Limit Comparison Test) If $A_n, B_n > 0$ and $lim A_n/B_n $ converges in $(0, infty)$, then $sum_n A_n$ converges if and only if $sum_n B_n$ converges.
$sum_{n=1}^{N} (log a_n - log a_{n-1}) = log a_N - log a_0 to infty$ as $Ntoinfty$, provided $lim a_n = infty$.
answered Jan 7 at 14:15
Sangchul LeeSangchul Lee
92.7k12167269
answered Jan 7 at 14:15
Sangchul LeeSangchul Lee
92.7k12167269
answered Jan 7 at 14:15
Sangchul LeeSangchul Lee
92.7k12167269
answered Jan 7 at 14:15
Sangchul LeeSangchul Lee
92.7k12167269
92.7k12167269
$begingroup$
A truly splendid argument, thank you!
$endgroup$
– Frederik vom Ende
Jan 7 at 14:27
1
$begingroup$
@FrederikvomEnde, Glad it helps! I realized that my argument assumes $b_n > 0$ for all $n$, but this causes no harm as we can remove repetitions from $(a_n)$, i.e., we have $$sum_{n=2}^{infty} left(1 - frac{a_{n-1}}{a_n} right) = sum_{k=1}^{infty} left(1 - frac{a_{n_{k-1}}}{a_{n_k}} right) $$ where $n = n_k$ is the $k$-th smallest index at which $a_{n-1} < a_n$ holds (with $n_0 := 1$).
$endgroup$
– Sangchul Lee
Jan 7 at 14:33
add a comment |
$begingroup$
A truly splendid argument, thank you!
$endgroup$
– Frederik vom Ende
Jan 7 at 14:27
1
$begingroup$
@FrederikvomEnde, Glad it helps! I realized that my argument assumes $b_n > 0$ for all $n$, but this causes no harm as we can remove repetitions from $(a_n)$, i.e., we have $$sum_{n=2}^{infty} left(1 - frac{a_{n-1}}{a_n} right) = sum_{k=1}^{infty} left(1 - frac{a_{n_{k-1}}}{a_{n_k}} right) $$ where $n = n_k$ is the $k$-th smallest index at which $a_{n-1} < a_n$ holds (with $n_0 := 1$).
$endgroup$
– Sangchul Lee
Jan 7 at 14:33
$begingroup$
A truly splendid argument, thank you!
$endgroup$
– Frederik vom Ende
Jan 7 at 14:27
$begingroup$
A truly splendid argument, thank you!
$endgroup$
– Frederik vom Ende
Jan 7 at 14:27
1
1
$begingroup$
@FrederikvomEnde, Glad it helps! I realized that my argument assumes $b_n > 0$ for all $n$, but this causes no harm as we can remove repetitions from $(a_n)$, i.e., we have $$sum_{n=2}^{infty} left(1 - frac{a_{n-1}}{a_n} right) = sum_{k=1}^{infty} left(1 - frac{a_{n_{k-1}}}{a_{n_k}} right) $$ where $n = n_k$ is the $k$-th smallest index at which $a_{n-1} < a_n$ holds (with $n_0 := 1$).
$endgroup$
– Sangchul Lee
Jan 7 at 14:33
$begingroup$
@FrederikvomEnde, Glad it helps! I realized that my argument assumes $b_n > 0$ for all $n$, but this causes no harm as we can remove repetitions from $(a_n)$, i.e., we have $$sum_{n=2}^{infty} left(1 - frac{a_{n-1}}{a_n} right) = sum_{k=1}^{infty} left(1 - frac{a_{n_{k-1}}}{a_{n_k}} right) $$ where $n = n_k$ is the $k$-th smallest index at which $a_{n-1} < a_n$ holds (with $n_0 := 1$).
$endgroup$
– Sangchul Lee
Jan 7 at 14:33
add a comment |
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$begingroup$
Simpler (?) proofs here math.stackexchange.com/q/746257/42969 and here math.stackexchange.com/q/388898/42969.
$endgroup$
– Martin R
Jan 7 at 13:31
$begingroup$
@Henry Changed the title to get rid of the ambiguous wording, thank you!
$endgroup$
– Frederik vom Ende
Jan 7 at 13:51
$begingroup$
@MartinR I'm not sure if the proof in question is directly applicable as the denominator differs from the one used here, i.e. $(a_{n+1}-a_n)/a_n$ vs. $(a_{n+1}-a_n)/a_{n+1}$ (with the latter being smaller so the above result should even be slightly stronger than the linked one?)
$endgroup$
– Frederik vom Ende
Jan 7 at 14:17
$begingroup$
Using lowercase letters for your sequence, and uppercase letters for the sequences in math.stackexchange.com/q/388898/42969, we have $a_n = S_n$, $a_n - a_{n-1} = S_n - S_{n-1} = A_n$, and therefore the identical series$ sum frac{a_n-a_{n-1}}{a_n} =sum frac{A_n}{S_n}$. Or did I make some error?
$endgroup$
– Martin R
Jan 7 at 14:22
$begingroup$
Yeah you are of course correct; my annotation rather was related to the post I linked in my comment (which seems to not be equivalent to, but implied by the above result). Anyways, thank you for your time and the useful links!
$endgroup$
– Frederik vom Ende
Jan 7 at 14:33