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I need to show that

$$\lim_{n\to\infty} \sum_{k=1}^n\frac{B_k}{k!}\,\frac{n^\underline{k-1}}{n^{k-1}}=\sum_{k=1}^\infty\frac{B_k}{k!}$$

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 $k\in\Bbb N_{\ge 1}$, 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

$$\left|\sum_{k=1}^\infty a_n(k)-\sum_{k=1}^\infty\frac{B_k}{k!}\right|\le\left|\sum_{k=1}^M(a_n(k)-B_k/k!)\right|+\sum_{k=M+1}^n\left|1-\frac{n^\underline{k-1}}{n^{k-1}}\right|+\left|\sum_{k=n+1}^\infty\frac{B_k}{k!}\right|$$

such that $|B_k/k!|<1$ for $k\ge M+1$. Then taking limits above we have that

$$\lim_{n\to\infty}\left|\sum_{k=1}^\infty\left(a_n(k)-\frac{B_k}{k!}\right)\right|\le\lim_{n\to\infty}\sum_{k=M+1}^n\left(1-\frac{n^\underline{k-1}}{n^{k-1}}\right)$$

for any fixed enough large $M$. Then if I can show that for each $\epsilon>0$ there is some $M\in\Bbb N$ such that

$$\lim_{n\to\infty}\sum_{k=M+1}^n\left(1-\frac{n^\underline{k-1}}{n^{k-1}}\right)<\epsilon$$

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.

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  • $\begingroup$ 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! $\endgroup$
    – p4sch
    Nov 19 '18 at 14:05
  • $\begingroup$ @p4sch I fixed it, thank you. However I find a solution! $\endgroup$
    – Masacroso
    Nov 19 '18 at 14:09
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    $\begingroup$ 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 $n$. $\endgroup$
    – p4sch
    Nov 19 '18 at 14:31
  • $\begingroup$ 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!$. $\endgroup$
    – p4sch
    Nov 19 '18 at 14:43
  • $\begingroup$ @p4sch yes, this was a persistent typographic error, thank you $\endgroup$
    – Masacroso
    Nov 19 '18 at 14:47
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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 $|\mathrm{Im}(z)| < 2\pi$. 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 $$\tag{1}\sum_{k=M+1}^n (1-n^{\underline{k-1}}/n^{k-1}) \ge \frac{1}{2} \sum_{k\ge n/2+2}^n \exp(-n^{\underline{k-1}}/n^{k-1}),$$ 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 $$ \frac{n}{2} \exp(- \exp\{-n/8\}) \rightarrow \infty.$$

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    $\begingroup$ $|B_k 2^k /k! | \le 1$ is nothing else than $|B_k/k!| \le 2^{-k}$. $\endgroup$
    – p4sch
    Nov 19 '18 at 14:20
  • $\begingroup$ 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 $n\to\infty$ for fixed $k$ $\endgroup$
    – Masacroso
    Nov 19 '18 at 21:29
  • $\begingroup$ $k$ 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 $k \ge n/2+2$. $\endgroup$
    – p4sch
    Nov 20 '18 at 8:26

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