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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
22 hours ago
@p4sch I fixed it, thank you. However I find a solution!
– Masacroso
22 hours ago
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
22 hours ago
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
21 hours ago
@p4sch yes, this was a persistent typographic error, thank you
– Masacroso
21 hours ago
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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 21 hours ago
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
22 hours ago
@p4sch I fixed it, thank you. However I find a solution!
– Masacroso
22 hours ago
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
22 hours ago
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
21 hours ago
@p4sch yes, this was a persistent typographic error, thank you
– Masacroso
21 hours ago
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
22 hours ago
@p4sch I fixed it, thank you. However I find a solution!
– Masacroso
22 hours ago
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
22 hours ago
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
21 hours ago
@p4sch yes, this was a persistent typographic error, thank you
– Masacroso
21 hours ago
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
22 hours ago
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
22 hours ago
@p4sch I fixed it, thank you. However I find a solution!
– Masacroso
22 hours ago
@p4sch I fixed it, thank you. However I find a solution!
– Masacroso
22 hours ago
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
22 hours ago
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
22 hours ago
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
21 hours ago
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
21 hours ago
@p4sch yes, this was a persistent typographic error, thank you
– Masacroso
21 hours ago
@p4sch yes, this was a persistent typographic error, thank you
– Masacroso
21 hours ago
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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 22 hours ago
p4sch
3,465216
1
$|B_k 2^k /k! | le 1$ is nothing else than $|B_k/k!| le 2^{-k}$.
– p4sch
22 hours ago
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
15 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
4 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 22 hours ago
Masacroso
12.2k41746
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2 Answers
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2 Answers
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active
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up vote
1
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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 22 hours ago
p4sch
3,465216
1
$|B_k 2^k /k! | le 1$ is nothing else than $|B_k/k!| le 2^{-k}$.
– p4sch
22 hours ago
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
15 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
4 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 22 hours ago
p4sch
3,465216
1
$|B_k 2^k /k! | le 1$ is nothing else than $|B_k/k!| le 2^{-k}$.
– p4sch
22 hours ago
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
15 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
4 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 22 hours ago
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 22 hours ago
p4sch
3,465216
edited 4 hours ago
answered 22 hours ago
p4sch
3,465216
answered 22 hours ago
p4sch
3,465216
answered 22 hours ago
p4sch
3,465216
3,465216
1
$|B_k 2^k /k! | le 1$ is nothing else than $|B_k/k!| le 2^{-k}$.
– p4sch
22 hours ago
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
15 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
4 hours ago
add a comment |
1
$|B_k 2^k /k! | le 1$ is nothing else than $|B_k/k!| le 2^{-k}$.
– p4sch
22 hours ago
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
15 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
4 hours ago
1
1
$|B_k 2^k /k! | le 1$ is nothing else than $|B_k/k!| le 2^{-k}$.
– p4sch
22 hours ago
$|B_k 2^k /k! | le 1$ is nothing else than $|B_k/k!| le 2^{-k}$.
– p4sch
22 hours ago
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
15 hours ago
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
15 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
4 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
4 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 22 hours ago
Masacroso
12.2k41746
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 22 hours ago
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 22 hours ago
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 22 hours ago
Masacroso
12.2k41746
edited 16 hours ago
answered 22 hours ago
Masacroso
12.2k41746
answered 22 hours ago
Masacroso
12.2k41746
answered 22 hours ago
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
22 hours ago
@p4sch I fixed it, thank you. However I find a solution!
– Masacroso
22 hours ago
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
22 hours ago
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
21 hours ago
@p4sch yes, this was a persistent typographic error, thank you
– Masacroso
21 hours ago