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Show that $lim_n sum_{k=1}^nfrac{B_k}{k!},frac{n^underline{k-1}}{n^{k-1}}=sum_{k=1}^inftyfrac{B_k}{k!}$
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I need to show that
$$lim_{ntoinfty} sum_{k=1}^nfrac{B_k}{k!},frac{n^underline{k-1}}{n^{k-1}}=sum_{k=1}^inftyfrac{B_k}{k!}$$
where $n^underline{k-1}:=prod_{j=0}^{k-2}(n-j)$ is a falling factorial and the $B_k$ are the Bernoulli numbers, and I know that the RHS of above converges to $1/(e-1)$. I had two attempts:
1) First I tried to use the dominated convergence theorem setting $a_n(k):=frac{B_k}{k!},frac{n^underline{k-1}}{n^{k-1}}chi_{[1,n]}(k)$, then clearly $lim_n a_n(k)=B_k/k!$ for each $kinBbb N_{ge 1}$, however I dont know if $sum_{k=1}^infty|B_k|/k!$ converges, and I dont know any absolutely convergent series that dominates, so Im stuck at this step.
2) A more elementary approach
$$left|sum_{k=1}^infty a_n(k)-sum_{k=1}^inftyfrac{B_k}{k!}right|leleft|sum_{k=1}^M(a_n(k)-B_k/k!)right|+sum_{k=M+1}^nleft|1-frac{n^underline{k-1}}{n^{k-1}}right|+left|sum_{k=n+1}^inftyfrac{B_k}{k!}right|$$
such that $|B_k/k!|<1$ for $kge M+1$. Then taking limits above we have that
$$lim_{ntoinfty}left|sum_{k=1}^inftyleft(a_n(k)-frac{B_k}{k!}right)right|lelim_{ntoinfty}sum_{k=M+1}^nleft(1-frac{n^underline{k-1}}{n^{k-1}}right)$$
for any fixed enough large $M$. Then if I can show that for each $epsilon>0$ there is some $MinBbb N$ such that
$$lim_{ntoinfty}sum_{k=M+1}^nleft(1-frac{n^underline{k-1}}{n^{k-1}}right)<epsilon$$
then Im done. However it is not clear how to accomplish (or if it is possible) this task. I thought about use the Stirling approximation on $n^underline{k-1}/n^{k-1}$, however it is not clear that I can apply an asymptotic expression inside a series, so Im again stuck.
There is some easy way (the more elementary the better) to show the converge of the limit of the title? Thank you.
real-analysis limits convergence
asked yesterday
Masacroso
12.2k41746
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I need to show that
$$lim_{ntoinfty} sum_{k=1}^nfrac{B_k}{k!},frac{n^underline{k-1}}{n^{k-1}}=sum_{k=1}^inftyfrac{B_k}{k!}$$
where $n^underline{k-1}:=prod_{j=0}^{k-2}(n-j)$ is a falling factorial and the $B_k$ are the Bernoulli numbers, and I know that the RHS of above converges to $1/(e-1)$. I had two attempts:
1) First I tried to use the dominated convergence theorem setting $a_n(k):=frac{B_k}{k!},frac{n^underline{k-1}}{n^{k-1}}chi_{[1,n]}(k)$, then clearly $lim_n a_n(k)=B_k/k!$ for each $kinBbb N_{ge 1}$, however I dont know if $sum_{k=1}^infty|B_k|/k!$ converges, and I dont know any absolutely convergent series that dominates, so Im stuck at this step.
2) A more elementary approach
$$left|sum_{k=1}^infty a_n(k)-sum_{k=1}^inftyfrac{B_k}{k!}right|leleft|sum_{k=1}^M(a_n(k)-B_k/k!)right|+sum_{k=M+1}^nleft|1-frac{n^underline{k-1}}{n^{k-1}}right|+left|sum_{k=n+1}^inftyfrac{B_k}{k!}right|$$
such that $|B_k/k!|<1$ for $kge M+1$. Then taking limits above we have that
$$lim_{ntoinfty}left|sum_{k=1}^inftyleft(a_n(k)-frac{B_k}{k!}right)right|lelim_{ntoinfty}sum_{k=M+1}^nleft(1-frac{n^underline{k-1}}{n^{k-1}}right)$$
for any fixed enough large $M$. Then if I can show that for each $epsilon>0$ there is some $MinBbb N$ such that
$$lim_{ntoinfty}sum_{k=M+1}^nleft(1-frac{n^underline{k-1}}{n^{k-1}}right)<epsilon$$
then Im done. However it is not clear how to accomplish (or if it is possible) this task. I thought about use the Stirling approximation on $n^underline{k-1}/n^{k-1}$, however it is not clear that I can apply an asymptotic expression inside a series, so Im again stuck.
There is some easy way (the more elementary the better) to show the converge of the limit of the title? Thank you.
real-analysis limits convergence
asked yesterday
Masacroso
12.2k41746
Since $n^{underline{k-1}} =0$ if $k-2 ge n$ the series $sum_{k=M+1}^infty (1-n^{underline{k-1}}/n^k)$ is divergent!
– p4sch
yesterday
@p4sch I fixed it, thank you. However I find a solution!
– Masacroso
yesterday
1
The same problem occurs in the last line: $M^{underline{k-1}} =0$ if $k ge M+2$. You would like to show that $sum_{k=M+1}^n (1-n^{underline{k-1}}/n^k)$ can be made small for all large $n ge N$ independent of $n$.
– p4sch
yesterday
There is also another problem in your post: Note that $n^{underline{k-1}}$ has only $k-1$ terms in the product. Thus $n^{underline{k-1}}/n^k = n^{-1} prod_{j=0}^{k-2} (1-j/n) le n^{-1}$. I.e. your limes is zero, but $ 0 > (2-e)/(e-1) =1/(e-1)-1 = sum_{k=1}^infty B_k/k!$.
– p4sch
yesterday
@p4sch yes, this was a persistent typographic error, thank you
– Masacroso
yesterday
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up vote
3
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up vote
3
down vote
favorite
I need to show that
$$lim_{ntoinfty} sum_{k=1}^nfrac{B_k}{k!},frac{n^underline{k-1}}{n^{k-1}}=sum_{k=1}^inftyfrac{B_k}{k!}$$
where $n^underline{k-1}:=prod_{j=0}^{k-2}(n-j)$ is a falling factorial and the $B_k$ are the Bernoulli numbers, and I know that the RHS of above converges to $1/(e-1)$. I had two attempts:
1) First I tried to use the dominated convergence theorem setting $a_n(k):=frac{B_k}{k!},frac{n^underline{k-1}}{n^{k-1}}chi_{[1,n]}(k)$, then clearly $lim_n a_n(k)=B_k/k!$ for each $kinBbb N_{ge 1}$, however I dont know if $sum_{k=1}^infty|B_k|/k!$ converges, and I dont know any absolutely convergent series that dominates, so Im stuck at this step.
2) A more elementary approach
$$left|sum_{k=1}^infty a_n(k)-sum_{k=1}^inftyfrac{B_k}{k!}right|leleft|sum_{k=1}^M(a_n(k)-B_k/k!)right|+sum_{k=M+1}^nleft|1-frac{n^underline{k-1}}{n^{k-1}}right|+left|sum_{k=n+1}^inftyfrac{B_k}{k!}right|$$
such that $|B_k/k!|<1$ for $kge M+1$. Then taking limits above we have that
$$lim_{ntoinfty}left|sum_{k=1}^inftyleft(a_n(k)-frac{B_k}{k!}right)right|lelim_{ntoinfty}sum_{k=M+1}^nleft(1-frac{n^underline{k-1}}{n^{k-1}}right)$$
for any fixed enough large $M$. Then if I can show that for each $epsilon>0$ there is some $MinBbb N$ such that
$$lim_{ntoinfty}sum_{k=M+1}^nleft(1-frac{n^underline{k-1}}{n^{k-1}}right)<epsilon$$
then Im done. However it is not clear how to accomplish (or if it is possible) this task. I thought about use the Stirling approximation on $n^underline{k-1}/n^{k-1}$, however it is not clear that I can apply an asymptotic expression inside a series, so Im again stuck.
There is some easy way (the more elementary the better) to show the converge of the limit of the title? Thank you.
real-analysis limits convergence
asked yesterday
Masacroso
12.2k41746
I need to show that
$$lim_{ntoinfty} sum_{k=1}^nfrac{B_k}{k!},frac{n^underline{k-1}}{n^{k-1}}=sum_{k=1}^inftyfrac{B_k}{k!}$$
where $n^underline{k-1}:=prod_{j=0}^{k-2}(n-j)$ is a falling factorial and the $B_k$ are the Bernoulli numbers, and I know that the RHS of above converges to $1/(e-1)$. I had two attempts:
1) First I tried to use the dominated convergence theorem setting $a_n(k):=frac{B_k}{k!},frac{n^underline{k-1}}{n^{k-1}}chi_{[1,n]}(k)$, then clearly $lim_n a_n(k)=B_k/k!$ for each $kinBbb N_{ge 1}$, however I dont know if $sum_{k=1}^infty|B_k|/k!$ converges, and I dont know any absolutely convergent series that dominates, so Im stuck at this step.
2) A more elementary approach
$$left|sum_{k=1}^infty a_n(k)-sum_{k=1}^inftyfrac{B_k}{k!}right|leleft|sum_{k=1}^M(a_n(k)-B_k/k!)right|+sum_{k=M+1}^nleft|1-frac{n^underline{k-1}}{n^{k-1}}right|+left|sum_{k=n+1}^inftyfrac{B_k}{k!}right|$$
such that $|B_k/k!|<1$ for $kge M+1$. Then taking limits above we have that
$$lim_{ntoinfty}left|sum_{k=1}^inftyleft(a_n(k)-frac{B_k}{k!}right)right|lelim_{ntoinfty}sum_{k=M+1}^nleft(1-frac{n^underline{k-1}}{n^{k-1}}right)$$
for any fixed enough large $M$. Then if I can show that for each $epsilon>0$ there is some $MinBbb N$ such that
$$lim_{ntoinfty}sum_{k=M+1}^nleft(1-frac{n^underline{k-1}}{n^{k-1}}right)<epsilon$$
then Im done. However it is not clear how to accomplish (or if it is possible) this task. I thought about use the Stirling approximation on $n^underline{k-1}/n^{k-1}$, however it is not clear that I can apply an asymptotic expression inside a series, so Im again stuck.
There is some easy way (the more elementary the better) to show the converge of the limit of the title? Thank you.
real-analysis limits convergence
real-analysis limits convergence
asked yesterday
Masacroso
12.2k41746
asked yesterday
Masacroso
12.2k41746
edited yesterday
asked yesterday
Masacroso
12.2k41746
asked yesterday
Masacroso
12.2k41746
asked yesterday
Masacroso
12.2k41746
12.2k41746
Since $n^{underline{k-1}} =0$ if $k-2 ge n$ the series $sum_{k=M+1}^infty (1-n^{underline{k-1}}/n^k)$ is divergent!
– p4sch
yesterday
@p4sch I fixed it, thank you. However I find a solution!
– Masacroso
yesterday
1
The same problem occurs in the last line: $M^{underline{k-1}} =0$ if $k ge M+2$. You would like to show that $sum_{k=M+1}^n (1-n^{underline{k-1}}/n^k)$ can be made small for all large $n ge N$ independent of $n$.
– p4sch
yesterday
There is also another problem in your post: Note that $n^{underline{k-1}}$ has only $k-1$ terms in the product. Thus $n^{underline{k-1}}/n^k = n^{-1} prod_{j=0}^{k-2} (1-j/n) le n^{-1}$. I.e. your limes is zero, but $ 0 > (2-e)/(e-1) =1/(e-1)-1 = sum_{k=1}^infty B_k/k!$.
– p4sch
yesterday
@p4sch yes, this was a persistent typographic error, thank you
– Masacroso
yesterday
add a comment |
Since $n^{underline{k-1}} =0$ if $k-2 ge n$ the series $sum_{k=M+1}^infty (1-n^{underline{k-1}}/n^k)$ is divergent!
– p4sch
yesterday
@p4sch I fixed it, thank you. However I find a solution!
– Masacroso
yesterday
1
The same problem occurs in the last line: $M^{underline{k-1}} =0$ if $k ge M+2$. You would like to show that $sum_{k=M+1}^n (1-n^{underline{k-1}}/n^k)$ can be made small for all large $n ge N$ independent of $n$.
– p4sch
yesterday
There is also another problem in your post: Note that $n^{underline{k-1}}$ has only $k-1$ terms in the product. Thus $n^{underline{k-1}}/n^k = n^{-1} prod_{j=0}^{k-2} (1-j/n) le n^{-1}$. I.e. your limes is zero, but $ 0 > (2-e)/(e-1) =1/(e-1)-1 = sum_{k=1}^infty B_k/k!$.
– p4sch
yesterday
@p4sch yes, this was a persistent typographic error, thank you
– Masacroso
yesterday
Since $n^{underline{k-1}} =0$ if $k-2 ge n$ the series $sum_{k=M+1}^infty (1-n^{underline{k-1}}/n^k)$ is divergent!
– p4sch
yesterday
Since $n^{underline{k-1}} =0$ if $k-2 ge n$ the series $sum_{k=M+1}^infty (1-n^{underline{k-1}}/n^k)$ is divergent!
– p4sch
yesterday
@p4sch I fixed it, thank you. However I find a solution!
– Masacroso
yesterday
@p4sch I fixed it, thank you. However I find a solution!
– Masacroso
yesterday
1
1
The same problem occurs in the last line: $M^{underline{k-1}} =0$ if $k ge M+2$. You would like to show that $sum_{k=M+1}^n (1-n^{underline{k-1}}/n^k)$ can be made small for all large $n ge N$ independent of $n$.
– p4sch
yesterday
The same problem occurs in the last line: $M^{underline{k-1}} =0$ if $k ge M+2$. You would like to show that $sum_{k=M+1}^n (1-n^{underline{k-1}}/n^k)$ can be made small for all large $n ge N$ independent of $n$.
– p4sch
yesterday
There is also another problem in your post: Note that $n^{underline{k-1}}$ has only $k-1$ terms in the product. Thus $n^{underline{k-1}}/n^k = n^{-1} prod_{j=0}^{k-2} (1-j/n) le n^{-1}$. I.e. your limes is zero, but $ 0 > (2-e)/(e-1) =1/(e-1)-1 = sum_{k=1}^infty B_k/k!$.
– p4sch
yesterday
There is also another problem in your post: Note that $n^{underline{k-1}}$ has only $k-1$ terms in the product. Thus $n^{underline{k-1}}/n^k = n^{-1} prod_{j=0}^{k-2} (1-j/n) le n^{-1}$. I.e. your limes is zero, but $ 0 > (2-e)/(e-1) =1/(e-1)-1 = sum_{k=1}^infty B_k/k!$.
– p4sch
yesterday
@p4sch yes, this was a persistent typographic error, thank you
– Masacroso
yesterday
@p4sch yes, this was a persistent typographic error, thank you
– Masacroso
yesterday
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The Bernoulli numbers are defined
$$frac{z}{e^z-1} = sum_{k=0}^infty frac{B_k}{k!}z^k$$
and this is a holomorphic function on the domain defined by $|mathrm{Im}(z)| < 2pi$. Thus the radius of convergence is $2 pi$. Any power seriers is absolutely convergent in the radius of convergence.
Here, for example we know that $|frac{B_k}{k!} 2^k| leq 1$ for all $k ge N$ and some $N in mathbb{N}$, because the series is convergent for $z=2$. Thus
$$sum_{k=N}^n frac{|B_k|}{k!} = sum_{k=N}^n frac{1}{2^k} le 1. $$
In other words: The series is absolute convergent and this can be used in order to show the claim (by using the dominated convergence theorem).
Second answer: The sum $sum_{k=M+1}^n (1-n^{underline{k-1}}/n^{k-1})$ isn't convergent, i.e. you have you use a better bound for $|B_k|/k!$ in order to get that the series is convergent. We use $1-x ge e^{-x}/2$ for $0 <x le 1/2$ to get
$$tag{1}sum_{k=M+1}^n (1-n^{underline{k-1}}/n^{k-1}) ge frac{1}{2} sum_{kge n/2+2}^n exp(-n^{underline{k-1}}/n^{k-1}),$$
where we used that
$$n^{underline{k-1}}/n^{k-1} = prod_{j=0}^{k-2} (1-j/n) le Big(1-frac{k-2}{n}Big)$$
is at most $1/2$ if $k ge n/2+2$. Now note that $ln(1-x) le -x$ for all $x ge 0$ to obtain also the lower bound
$$ ln (n^{underline{k-1}}/n^{k-1}) = sum_{j=0}^{k-2} ln(1-j/n) le -sum_{j=1}^{k-2} j/n = - frac{(k-1)(k-2)}{2n}.$$
All in all, we see for $n ge 2M-4$ that (1) can be bounded below by
$$frac{1}{2} sum_{k=lfloor n/2+2 rfloor +1}^n exp[-exp {-(k-1)(k-2)/(2n)}].$$
and this is bounded by
$$ frac{n}{2} exp(- exp{-n/8}) rightarrow infty.$$
answered yesterday
p4sch
3,465216
1
$|B_k 2^k /k! | le 1$ is nothing else than $|B_k/k!| le 2^{-k}$.
– p4sch
yesterday
in your second answer you are saying that $|n^underline{k-1}/n^{k-1}|<1/2$, but this is relative to the value of $k$ and $n$, so you cant make the lower bound on $(1)$ because $n^underline{k-1}/n^{k-1}to 1$ as $ntoinfty$ for fixed $k$
– Masacroso
yesterday
$k$ is not fixed, since we sum over $k=M+1$ up to $k=n$. I repaired my answer. You just need to skip sum summands and begin to sum at $k ge n/2+2$.
– p4sch
16 hours ago
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I find a solution using the recursion
$$B_k=-sum_{j=0}^{k-1}binom{k}jfrac{B_j}{k-1+j}tag1$$
and the dominated convergence theorem. From $(1)$ and setting $M:=sup_{kinBbb N}|B_k|/k!$ (what is finite because I know that $sum_{k=1}^infty B_k/k!$ converges) then we have that
$$begin{align}left|frac{B_k}{k!}right|&=left|frac1{k!}sum_{j=0}^{k-1}binom{k}jfrac{B_j}{k-j+1}right|=left|sum_{j=0}^{k-1}frac{B_j}{j!}frac1{(k-j+1)!}right|\
&lefrac{M k}{(k-1)!}sum_{j=0}^{k-1}binom{k-1}j
=frac{Mk}{(k-1)!}2^{k-1}\&=2kMprod_{j=0}^{k-4}frac2{k-1-j}end{align}tag2$$
Then it is easy to check that for $kge 6$ we have that $frac{|B_k|}{k!}le 16Mfrac{k}{(k-3)^3}$, so by the integral test the series $sum_{k=1}^infty |B_k|/k!$ converges.
Now using the first approach on the question we can apply the dominated convergence theorem and find that
$$lim_{ntoinfty}sum_{k=1}^{n+1}frac{B_k}{k!}cdotfrac{n^underline{k-1}}{n^{k-1}}=lim_{ntoinfty}int_{Bbb N_{ge 1}}a_n(k) ,dmathcal H^0=int_{Bbb N_{ge 1}}frac{B_k}{k!},dmathcal H^0=sum_{k=1}^inftyfrac{B_k}{k!}tag3$$
as desired. However I would like to see, if possible, a more simple solution that doesn't use "advanced" theorems as the dominated convergence theorem.
P.S.: $dmathcal H^0$ is just the counting measure.
Essentially the same question appear in a mathematical magazine. In short it says that $n^underline{k-1}in n^{k-1}+O(n^{k-2})$ as $ntoinfty$, thus $frac{n^underline{k-1}}{n^{k-1}}in 1+O(n^{-1})$ and we find that
$$sum_{k=1}^{n+1}frac{B_k}{k!}cdotfrac{n^underline{k-1}}{n^{k-1}}in sum_{k=1}^{n+1}frac{B_k}{k!}+O(n^{-1})sum_{k=1}^{n+1}frac{B_k}{k!}tag4$$
Then in the article the author take limits on $(4)$, "giving" the wanted identity. However this procedure seems wrong as its said in the last link. I added this to the answer for the curious reader.
answered yesterday
Masacroso
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2 Answers
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2 Answers
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The Bernoulli numbers are defined
$$frac{z}{e^z-1} = sum_{k=0}^infty frac{B_k}{k!}z^k$$
and this is a holomorphic function on the domain defined by $|mathrm{Im}(z)| < 2pi$. Thus the radius of convergence is $2 pi$. Any power seriers is absolutely convergent in the radius of convergence.
Here, for example we know that $|frac{B_k}{k!} 2^k| leq 1$ for all $k ge N$ and some $N in mathbb{N}$, because the series is convergent for $z=2$. Thus
$$sum_{k=N}^n frac{|B_k|}{k!} = sum_{k=N}^n frac{1}{2^k} le 1. $$
In other words: The series is absolute convergent and this can be used in order to show the claim (by using the dominated convergence theorem).
Second answer: The sum $sum_{k=M+1}^n (1-n^{underline{k-1}}/n^{k-1})$ isn't convergent, i.e. you have you use a better bound for $|B_k|/k!$ in order to get that the series is convergent. We use $1-x ge e^{-x}/2$ for $0 <x le 1/2$ to get
$$tag{1}sum_{k=M+1}^n (1-n^{underline{k-1}}/n^{k-1}) ge frac{1}{2} sum_{kge n/2+2}^n exp(-n^{underline{k-1}}/n^{k-1}),$$
where we used that
$$n^{underline{k-1}}/n^{k-1} = prod_{j=0}^{k-2} (1-j/n) le Big(1-frac{k-2}{n}Big)$$
is at most $1/2$ if $k ge n/2+2$. Now note that $ln(1-x) le -x$ for all $x ge 0$ to obtain also the lower bound
$$ ln (n^{underline{k-1}}/n^{k-1}) = sum_{j=0}^{k-2} ln(1-j/n) le -sum_{j=1}^{k-2} j/n = - frac{(k-1)(k-2)}{2n}.$$
All in all, we see for $n ge 2M-4$ that (1) can be bounded below by
$$frac{1}{2} sum_{k=lfloor n/2+2 rfloor +1}^n exp[-exp {-(k-1)(k-2)/(2n)}].$$
and this is bounded by
$$ frac{n}{2} exp(- exp{-n/8}) rightarrow infty.$$
answered yesterday
p4sch
3,465216
1
$|B_k 2^k /k! | le 1$ is nothing else than $|B_k/k!| le 2^{-k}$.
– p4sch
yesterday
in your second answer you are saying that $|n^underline{k-1}/n^{k-1}|<1/2$, but this is relative to the value of $k$ and $n$, so you cant make the lower bound on $(1)$ because $n^underline{k-1}/n^{k-1}to 1$ as $ntoinfty$ for fixed $k$
– Masacroso
yesterday
$k$ is not fixed, since we sum over $k=M+1$ up to $k=n$. I repaired my answer. You just need to skip sum summands and begin to sum at $k ge n/2+2$.
– p4sch
16 hours ago
add a comment |
up vote
1
down vote
The Bernoulli numbers are defined
$$frac{z}{e^z-1} = sum_{k=0}^infty frac{B_k}{k!}z^k$$
and this is a holomorphic function on the domain defined by $|mathrm{Im}(z)| < 2pi$. Thus the radius of convergence is $2 pi$. Any power seriers is absolutely convergent in the radius of convergence.
Here, for example we know that $|frac{B_k}{k!} 2^k| leq 1$ for all $k ge N$ and some $N in mathbb{N}$, because the series is convergent for $z=2$. Thus
$$sum_{k=N}^n frac{|B_k|}{k!} = sum_{k=N}^n frac{1}{2^k} le 1. $$
In other words: The series is absolute convergent and this can be used in order to show the claim (by using the dominated convergence theorem).
Second answer: The sum $sum_{k=M+1}^n (1-n^{underline{k-1}}/n^{k-1})$ isn't convergent, i.e. you have you use a better bound for $|B_k|/k!$ in order to get that the series is convergent. We use $1-x ge e^{-x}/2$ for $0 <x le 1/2$ to get
$$tag{1}sum_{k=M+1}^n (1-n^{underline{k-1}}/n^{k-1}) ge frac{1}{2} sum_{kge n/2+2}^n exp(-n^{underline{k-1}}/n^{k-1}),$$
where we used that
$$n^{underline{k-1}}/n^{k-1} = prod_{j=0}^{k-2} (1-j/n) le Big(1-frac{k-2}{n}Big)$$
is at most $1/2$ if $k ge n/2+2$. Now note that $ln(1-x) le -x$ for all $x ge 0$ to obtain also the lower bound
$$ ln (n^{underline{k-1}}/n^{k-1}) = sum_{j=0}^{k-2} ln(1-j/n) le -sum_{j=1}^{k-2} j/n = - frac{(k-1)(k-2)}{2n}.$$
All in all, we see for $n ge 2M-4$ that (1) can be bounded below by
$$frac{1}{2} sum_{k=lfloor n/2+2 rfloor +1}^n exp[-exp {-(k-1)(k-2)/(2n)}].$$
and this is bounded by
$$ frac{n}{2} exp(- exp{-n/8}) rightarrow infty.$$
answered yesterday
p4sch
3,465216
1
$|B_k 2^k /k! | le 1$ is nothing else than $|B_k/k!| le 2^{-k}$.
– p4sch
yesterday
in your second answer you are saying that $|n^underline{k-1}/n^{k-1}|<1/2$, but this is relative to the value of $k$ and $n$, so you cant make the lower bound on $(1)$ because $n^underline{k-1}/n^{k-1}to 1$ as $ntoinfty$ for fixed $k$
– Masacroso
yesterday
$k$ is not fixed, since we sum over $k=M+1$ up to $k=n$. I repaired my answer. You just need to skip sum summands and begin to sum at $k ge n/2+2$.
– p4sch
16 hours ago
add a comment |
up vote
1
down vote
up vote
1
down vote
The Bernoulli numbers are defined
$$frac{z}{e^z-1} = sum_{k=0}^infty frac{B_k}{k!}z^k$$
and this is a holomorphic function on the domain defined by $|mathrm{Im}(z)| < 2pi$. Thus the radius of convergence is $2 pi$. Any power seriers is absolutely convergent in the radius of convergence.
Here, for example we know that $|frac{B_k}{k!} 2^k| leq 1$ for all $k ge N$ and some $N in mathbb{N}$, because the series is convergent for $z=2$. Thus
$$sum_{k=N}^n frac{|B_k|}{k!} = sum_{k=N}^n frac{1}{2^k} le 1. $$
In other words: The series is absolute convergent and this can be used in order to show the claim (by using the dominated convergence theorem).
Second answer: The sum $sum_{k=M+1}^n (1-n^{underline{k-1}}/n^{k-1})$ isn't convergent, i.e. you have you use a better bound for $|B_k|/k!$ in order to get that the series is convergent. We use $1-x ge e^{-x}/2$ for $0 <x le 1/2$ to get
$$tag{1}sum_{k=M+1}^n (1-n^{underline{k-1}}/n^{k-1}) ge frac{1}{2} sum_{kge n/2+2}^n exp(-n^{underline{k-1}}/n^{k-1}),$$
where we used that
$$n^{underline{k-1}}/n^{k-1} = prod_{j=0}^{k-2} (1-j/n) le Big(1-frac{k-2}{n}Big)$$
is at most $1/2$ if $k ge n/2+2$. Now note that $ln(1-x) le -x$ for all $x ge 0$ to obtain also the lower bound
$$ ln (n^{underline{k-1}}/n^{k-1}) = sum_{j=0}^{k-2} ln(1-j/n) le -sum_{j=1}^{k-2} j/n = - frac{(k-1)(k-2)}{2n}.$$
All in all, we see for $n ge 2M-4$ that (1) can be bounded below by
$$frac{1}{2} sum_{k=lfloor n/2+2 rfloor +1}^n exp[-exp {-(k-1)(k-2)/(2n)}].$$
and this is bounded by
$$ frac{n}{2} exp(- exp{-n/8}) rightarrow infty.$$
answered yesterday
p4sch
3,465216
The Bernoulli numbers are defined
$$frac{z}{e^z-1} = sum_{k=0}^infty frac{B_k}{k!}z^k$$
and this is a holomorphic function on the domain defined by $|mathrm{Im}(z)| < 2pi$. Thus the radius of convergence is $2 pi$. Any power seriers is absolutely convergent in the radius of convergence.
Here, for example we know that $|frac{B_k}{k!} 2^k| leq 1$ for all $k ge N$ and some $N in mathbb{N}$, because the series is convergent for $z=2$. Thus
$$sum_{k=N}^n frac{|B_k|}{k!} = sum_{k=N}^n frac{1}{2^k} le 1. $$
In other words: The series is absolute convergent and this can be used in order to show the claim (by using the dominated convergence theorem).
Second answer: The sum $sum_{k=M+1}^n (1-n^{underline{k-1}}/n^{k-1})$ isn't convergent, i.e. you have you use a better bound for $|B_k|/k!$ in order to get that the series is convergent. We use $1-x ge e^{-x}/2$ for $0 <x le 1/2$ to get
$$tag{1}sum_{k=M+1}^n (1-n^{underline{k-1}}/n^{k-1}) ge frac{1}{2} sum_{kge n/2+2}^n exp(-n^{underline{k-1}}/n^{k-1}),$$
where we used that
$$n^{underline{k-1}}/n^{k-1} = prod_{j=0}^{k-2} (1-j/n) le Big(1-frac{k-2}{n}Big)$$
is at most $1/2$ if $k ge n/2+2$. Now note that $ln(1-x) le -x$ for all $x ge 0$ to obtain also the lower bound
$$ ln (n^{underline{k-1}}/n^{k-1}) = sum_{j=0}^{k-2} ln(1-j/n) le -sum_{j=1}^{k-2} j/n = - frac{(k-1)(k-2)}{2n}.$$
All in all, we see for $n ge 2M-4$ that (1) can be bounded below by
$$frac{1}{2} sum_{k=lfloor n/2+2 rfloor +1}^n exp[-exp {-(k-1)(k-2)/(2n)}].$$
and this is bounded by
$$ frac{n}{2} exp(- exp{-n/8}) rightarrow infty.$$
answered yesterday
p4sch
3,465216
edited 16 hours ago
answered yesterday
p4sch
3,465216
answered yesterday
p4sch
3,465216
answered yesterday
p4sch
3,465216
3,465216
1
$|B_k 2^k /k! | le 1$ is nothing else than $|B_k/k!| le 2^{-k}$.
– p4sch
yesterday
in your second answer you are saying that $|n^underline{k-1}/n^{k-1}|<1/2$, but this is relative to the value of $k$ and $n$, so you cant make the lower bound on $(1)$ because $n^underline{k-1}/n^{k-1}to 1$ as $ntoinfty$ for fixed $k$
– Masacroso
yesterday
$k$ is not fixed, since we sum over $k=M+1$ up to $k=n$. I repaired my answer. You just need to skip sum summands and begin to sum at $k ge n/2+2$.
– p4sch
16 hours ago
add a comment |
1
$|B_k 2^k /k! | le 1$ is nothing else than $|B_k/k!| le 2^{-k}$.
– p4sch
yesterday
in your second answer you are saying that $|n^underline{k-1}/n^{k-1}|<1/2$, but this is relative to the value of $k$ and $n$, so you cant make the lower bound on $(1)$ because $n^underline{k-1}/n^{k-1}to 1$ as $ntoinfty$ for fixed $k$
– Masacroso
yesterday
$k$ is not fixed, since we sum over $k=M+1$ up to $k=n$. I repaired my answer. You just need to skip sum summands and begin to sum at $k ge n/2+2$.
– p4sch
16 hours ago
1
1
$|B_k 2^k /k! | le 1$ is nothing else than $|B_k/k!| le 2^{-k}$.
– p4sch
yesterday
$|B_k 2^k /k! | le 1$ is nothing else than $|B_k/k!| le 2^{-k}$.
– p4sch
yesterday
in your second answer you are saying that $|n^underline{k-1}/n^{k-1}|<1/2$, but this is relative to the value of $k$ and $n$, so you cant make the lower bound on $(1)$ because $n^underline{k-1}/n^{k-1}to 1$ as $ntoinfty$ for fixed $k$
– Masacroso
yesterday
in your second answer you are saying that $|n^underline{k-1}/n^{k-1}|<1/2$, but this is relative to the value of $k$ and $n$, so you cant make the lower bound on $(1)$ because $n^underline{k-1}/n^{k-1}to 1$ as $ntoinfty$ for fixed $k$
– Masacroso
yesterday
$k$ is not fixed, since we sum over $k=M+1$ up to $k=n$. I repaired my answer. You just need to skip sum summands and begin to sum at $k ge n/2+2$.
– p4sch
16 hours ago
$k$ is not fixed, since we sum over $k=M+1$ up to $k=n$. I repaired my answer. You just need to skip sum summands and begin to sum at $k ge n/2+2$.
– p4sch
16 hours ago
add a comment |
up vote
0
down vote
I find a solution using the recursion
$$B_k=-sum_{j=0}^{k-1}binom{k}jfrac{B_j}{k-1+j}tag1$$
and the dominated convergence theorem. From $(1)$ and setting $M:=sup_{kinBbb N}|B_k|/k!$ (what is finite because I know that $sum_{k=1}^infty B_k/k!$ converges) then we have that
$$begin{align}left|frac{B_k}{k!}right|&=left|frac1{k!}sum_{j=0}^{k-1}binom{k}jfrac{B_j}{k-j+1}right|=left|sum_{j=0}^{k-1}frac{B_j}{j!}frac1{(k-j+1)!}right|\
&lefrac{M k}{(k-1)!}sum_{j=0}^{k-1}binom{k-1}j
=frac{Mk}{(k-1)!}2^{k-1}\&=2kMprod_{j=0}^{k-4}frac2{k-1-j}end{align}tag2$$
Then it is easy to check that for $kge 6$ we have that $frac{|B_k|}{k!}le 16Mfrac{k}{(k-3)^3}$, so by the integral test the series $sum_{k=1}^infty |B_k|/k!$ converges.
Now using the first approach on the question we can apply the dominated convergence theorem and find that
$$lim_{ntoinfty}sum_{k=1}^{n+1}frac{B_k}{k!}cdotfrac{n^underline{k-1}}{n^{k-1}}=lim_{ntoinfty}int_{Bbb N_{ge 1}}a_n(k) ,dmathcal H^0=int_{Bbb N_{ge 1}}frac{B_k}{k!},dmathcal H^0=sum_{k=1}^inftyfrac{B_k}{k!}tag3$$
as desired. However I would like to see, if possible, a more simple solution that doesn't use "advanced" theorems as the dominated convergence theorem.
P.S.: $dmathcal H^0$ is just the counting measure.
Essentially the same question appear in a mathematical magazine. In short it says that $n^underline{k-1}in n^{k-1}+O(n^{k-2})$ as $ntoinfty$, thus $frac{n^underline{k-1}}{n^{k-1}}in 1+O(n^{-1})$ and we find that
$$sum_{k=1}^{n+1}frac{B_k}{k!}cdotfrac{n^underline{k-1}}{n^{k-1}}in sum_{k=1}^{n+1}frac{B_k}{k!}+O(n^{-1})sum_{k=1}^{n+1}frac{B_k}{k!}tag4$$
Then in the article the author take limits on $(4)$, "giving" the wanted identity. However this procedure seems wrong as its said in the last link. I added this to the answer for the curious reader.
answered yesterday
Masacroso
12.2k41746
add a comment |
up vote
0
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I find a solution using the recursion
$$B_k=-sum_{j=0}^{k-1}binom{k}jfrac{B_j}{k-1+j}tag1$$
and the dominated convergence theorem. From $(1)$ and setting $M:=sup_{kinBbb N}|B_k|/k!$ (what is finite because I know that $sum_{k=1}^infty B_k/k!$ converges) then we have that
$$begin{align}left|frac{B_k}{k!}right|&=left|frac1{k!}sum_{j=0}^{k-1}binom{k}jfrac{B_j}{k-j+1}right|=left|sum_{j=0}^{k-1}frac{B_j}{j!}frac1{(k-j+1)!}right|\
&lefrac{M k}{(k-1)!}sum_{j=0}^{k-1}binom{k-1}j
=frac{Mk}{(k-1)!}2^{k-1}\&=2kMprod_{j=0}^{k-4}frac2{k-1-j}end{align}tag2$$
Then it is easy to check that for $kge 6$ we have that $frac{|B_k|}{k!}le 16Mfrac{k}{(k-3)^3}$, so by the integral test the series $sum_{k=1}^infty |B_k|/k!$ converges.
Now using the first approach on the question we can apply the dominated convergence theorem and find that
$$lim_{ntoinfty}sum_{k=1}^{n+1}frac{B_k}{k!}cdotfrac{n^underline{k-1}}{n^{k-1}}=lim_{ntoinfty}int_{Bbb N_{ge 1}}a_n(k) ,dmathcal H^0=int_{Bbb N_{ge 1}}frac{B_k}{k!},dmathcal H^0=sum_{k=1}^inftyfrac{B_k}{k!}tag3$$
as desired. However I would like to see, if possible, a more simple solution that doesn't use "advanced" theorems as the dominated convergence theorem.
P.S.: $dmathcal H^0$ is just the counting measure.
Essentially the same question appear in a mathematical magazine. In short it says that $n^underline{k-1}in n^{k-1}+O(n^{k-2})$ as $ntoinfty$, thus $frac{n^underline{k-1}}{n^{k-1}}in 1+O(n^{-1})$ and we find that
$$sum_{k=1}^{n+1}frac{B_k}{k!}cdotfrac{n^underline{k-1}}{n^{k-1}}in sum_{k=1}^{n+1}frac{B_k}{k!}+O(n^{-1})sum_{k=1}^{n+1}frac{B_k}{k!}tag4$$
Then in the article the author take limits on $(4)$, "giving" the wanted identity. However this procedure seems wrong as its said in the last link. I added this to the answer for the curious reader.
answered yesterday
Masacroso
12.2k41746
add a comment |
up vote
0
down vote
up vote
0
down vote
I find a solution using the recursion
$$B_k=-sum_{j=0}^{k-1}binom{k}jfrac{B_j}{k-1+j}tag1$$
and the dominated convergence theorem. From $(1)$ and setting $M:=sup_{kinBbb N}|B_k|/k!$ (what is finite because I know that $sum_{k=1}^infty B_k/k!$ converges) then we have that
$$begin{align}left|frac{B_k}{k!}right|&=left|frac1{k!}sum_{j=0}^{k-1}binom{k}jfrac{B_j}{k-j+1}right|=left|sum_{j=0}^{k-1}frac{B_j}{j!}frac1{(k-j+1)!}right|\
&lefrac{M k}{(k-1)!}sum_{j=0}^{k-1}binom{k-1}j
=frac{Mk}{(k-1)!}2^{k-1}\&=2kMprod_{j=0}^{k-4}frac2{k-1-j}end{align}tag2$$
Then it is easy to check that for $kge 6$ we have that $frac{|B_k|}{k!}le 16Mfrac{k}{(k-3)^3}$, so by the integral test the series $sum_{k=1}^infty |B_k|/k!$ converges.
Now using the first approach on the question we can apply the dominated convergence theorem and find that
$$lim_{ntoinfty}sum_{k=1}^{n+1}frac{B_k}{k!}cdotfrac{n^underline{k-1}}{n^{k-1}}=lim_{ntoinfty}int_{Bbb N_{ge 1}}a_n(k) ,dmathcal H^0=int_{Bbb N_{ge 1}}frac{B_k}{k!},dmathcal H^0=sum_{k=1}^inftyfrac{B_k}{k!}tag3$$
as desired. However I would like to see, if possible, a more simple solution that doesn't use "advanced" theorems as the dominated convergence theorem.
P.S.: $dmathcal H^0$ is just the counting measure.
Essentially the same question appear in a mathematical magazine. In short it says that $n^underline{k-1}in n^{k-1}+O(n^{k-2})$ as $ntoinfty$, thus $frac{n^underline{k-1}}{n^{k-1}}in 1+O(n^{-1})$ and we find that
$$sum_{k=1}^{n+1}frac{B_k}{k!}cdotfrac{n^underline{k-1}}{n^{k-1}}in sum_{k=1}^{n+1}frac{B_k}{k!}+O(n^{-1})sum_{k=1}^{n+1}frac{B_k}{k!}tag4$$
Then in the article the author take limits on $(4)$, "giving" the wanted identity. However this procedure seems wrong as its said in the last link. I added this to the answer for the curious reader.
answered yesterday
Masacroso
12.2k41746
I find a solution using the recursion
$$B_k=-sum_{j=0}^{k-1}binom{k}jfrac{B_j}{k-1+j}tag1$$
and the dominated convergence theorem. From $(1)$ and setting $M:=sup_{kinBbb N}|B_k|/k!$ (what is finite because I know that $sum_{k=1}^infty B_k/k!$ converges) then we have that
$$begin{align}left|frac{B_k}{k!}right|&=left|frac1{k!}sum_{j=0}^{k-1}binom{k}jfrac{B_j}{k-j+1}right|=left|sum_{j=0}^{k-1}frac{B_j}{j!}frac1{(k-j+1)!}right|\
&lefrac{M k}{(k-1)!}sum_{j=0}^{k-1}binom{k-1}j
=frac{Mk}{(k-1)!}2^{k-1}\&=2kMprod_{j=0}^{k-4}frac2{k-1-j}end{align}tag2$$
Then it is easy to check that for $kge 6$ we have that $frac{|B_k|}{k!}le 16Mfrac{k}{(k-3)^3}$, so by the integral test the series $sum_{k=1}^infty |B_k|/k!$ converges.
Now using the first approach on the question we can apply the dominated convergence theorem and find that
$$lim_{ntoinfty}sum_{k=1}^{n+1}frac{B_k}{k!}cdotfrac{n^underline{k-1}}{n^{k-1}}=lim_{ntoinfty}int_{Bbb N_{ge 1}}a_n(k) ,dmathcal H^0=int_{Bbb N_{ge 1}}frac{B_k}{k!},dmathcal H^0=sum_{k=1}^inftyfrac{B_k}{k!}tag3$$
as desired. However I would like to see, if possible, a more simple solution that doesn't use "advanced" theorems as the dominated convergence theorem.
P.S.: $dmathcal H^0$ is just the counting measure.
Essentially the same question appear in a mathematical magazine. In short it says that $n^underline{k-1}in n^{k-1}+O(n^{k-2})$ as $ntoinfty$, thus $frac{n^underline{k-1}}{n^{k-1}}in 1+O(n^{-1})$ and we find that
$$sum_{k=1}^{n+1}frac{B_k}{k!}cdotfrac{n^underline{k-1}}{n^{k-1}}in sum_{k=1}^{n+1}frac{B_k}{k!}+O(n^{-1})sum_{k=1}^{n+1}frac{B_k}{k!}tag4$$
Then in the article the author take limits on $(4)$, "giving" the wanted identity. However this procedure seems wrong as its said in the last link. I added this to the answer for the curious reader.
answered yesterday
Masacroso
12.2k41746
edited yesterday
answered yesterday
Masacroso
12.2k41746
answered yesterday
Masacroso
12.2k41746
answered yesterday
Masacroso
12.2k41746
12.2k41746
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Since $n^{underline{k-1}} =0$ if $k-2 ge n$ the series $sum_{k=M+1}^infty (1-n^{underline{k-1}}/n^k)$ is divergent!
– p4sch
yesterday
@p4sch I fixed it, thank you. However I find a solution!
– Masacroso
yesterday
1
The same problem occurs in the last line: $M^{underline{k-1}} =0$ if $k ge M+2$. You would like to show that $sum_{k=M+1}^n (1-n^{underline{k-1}}/n^k)$ can be made small for all large $n ge N$ independent of $n$.
– p4sch
yesterday
There is also another problem in your post: Note that $n^{underline{k-1}}$ has only $k-1$ terms in the product. Thus $n^{underline{k-1}}/n^k = n^{-1} prod_{j=0}^{k-2} (1-j/n) le n^{-1}$. I.e. your limes is zero, but $ 0 > (2-e)/(e-1) =1/(e-1)-1 = sum_{k=1}^infty B_k/k!$.
– p4sch
yesterday
@p4sch yes, this was a persistent typographic error, thank you
– Masacroso
yesterday