# Show that $\lim_n \sum_{k=1}^n\frac{B_k}{k!}\,\frac{n^\underline{k-1}}{n^{k-1}}=\sum_{k=1}^\infty\frac{B_k}{k!}$

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.

• 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! Nov 19 '18 at 14:05
• @p4sch I fixed it, thank you. However I find a solution! Nov 19 '18 at 14:09
• 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 $n$. Nov 19 '18 at 14:31
• 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 $0 > (2-e)/(e-1) =1/(e-1)-1 = \sum_{k=1}^\infty B_k/k!$. Nov 19 '18 at 14:43
• @p4sch yes, this was a persistent typographic error, thank you Nov 19 '18 at 14:47

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 $$0 to get $$\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.$$
• $|B_k 2^k /k! | \le 1$ is nothing else than $|B_k/k!| \le 2^{-k}$. Nov 19 '18 at 14:20
• 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 $k$ Nov 19 '18 at 21:29
• $k$ 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 $k \ge n/2+2$. Nov 20 '18 at 8:26