Mathcenter Forum - Bernoulli number

อ้างอิง:

Problem. Let $B_1,B_2,B_3,\ldots$ be the real numbers such that $$\frac{x}{\mathrm{e}^x-1}=1-\frac{x}{2}+\frac{B_1}{2!}\,x^2-\frac{B_2}{4!}\,x^4+\frac{B_3}{6!}\,x^6-\ldots\,.$$ Show that $$1^p+2^p+3^p+\ldots+n^p=\frac{n^{p+1}}{p+1}+\frac{n^p}{2}+\frac{B_1}{2!}\,p\,n^{p-1}-\frac{B_2}{4!}\,p(p-1)(p-2)\,n^{p-3}+\ldots\,.$$

Let $S^p_n$ denote the sum $1^p+2^p+3^p+\ldots+n^p$ for each positive integer $n$ and for each nonnegative integer $p$. Observe that
$$f_n(x):=\sum_{p=0}^\infty\,\frac{S^p_n}{p!}\,x^p=\sum_{p=0}^\infty\,\sum_{k=0}^n\,\frac{k^p}{p!}\,x^p=\sum_{k=1}^n\,\sum_{p=0} ^\infty\,\frac{(kx)^p}{p!}\,.$$
Therefore,
$$f_n(x)=\sum_{k=1}^n\,\exp(kx)=\frac{\big(\exp(nx)-1\big)\,\exp(x)}{\exp(x)-1}\,.$$
Consequently,
$$f_n(x)=\left(\frac{\exp(nx)-1}{x}\right)\,\left(\frac{x}{\exp(x)-1}\right)\,\exp(x)\,.$$
Now, we have
$$\exp(x)=\sum_{k=0}^\infty\,\frac{1}{k!}\,x^k\,,$$
$$\frac{\exp(nx)-1}{x}=\sum_{r=0}^\infty\,\frac{n^{r+1}}{(r+1)!}\,x^r\,,$$
and
$$\frac{x}{\exp(x)-1}=-\frac{x}{2}-\sum_{s=0}^\infty\,\frac{(-1)^s}{(2s)!}\,B_{s}\,x^{2s}\,,$$
where $B_0,B_1,B_2,\ldots$ are the Bernoulli numbers (with $B_0:=-1$). Thus, the coefficient of $x^p$ in $f_n(x)$ is given by the sum
$$s_n^p:=-\frac{1}{2}\,\sum_{r=0}^{p-1}\,\frac{n^{r+1}}{(r+1)!\,(p-1-r)!}+\sum_{\substack{r,s\geq 0\\ r+2s\leq p}}\,\frac{(-1)^s\,B_{s}\,n^{r+1}}{(r+1)!\,(2s)!\,(p-r-2s)!}\,.$$
That is,
$$\begin{align}S_n^p&=p!\,s_n^p\\&=\sum_{\ell=0}^{p+1}\,\frac{p!}{(p-\ell+1)!}\,n^{p+1-\ell}\,\left(-\sum_{s=0}^{\left\lfloor \frac{\ell}{2}\right\rfloor}\,\frac{(-1)^s\,B_s}{(2s)!\,(\ell-2s)!}-\frac{1}{2\,(\ell-1)!}\right)\,.\tag{$\star$}\end{align}$$
We need to verify that
$$-\sum_{s=0}^{\left\lfloor \frac{\ell}{2}\right\rfloor}\,(-1)^s\,\binom{\ell}{2s}\,B_s-\frac{\ell}{2}=\begin{cases}0&\text{if }\ell\geq 3\text{ is odd}\,,\\
-(-1)^{\frac{\ell}{2}}\,B_{\frac{\ell}{2}}&\text{if }\ell\geq 2\text{ is even}\,.
\end{cases}\tag{*}$$
For convenience, write
$$b_s:=\begin{cases}
-(-1)^{\frac{s}{2}}B_{\frac{s}{2}}&\text{if }s\geq 0\text{ is even}\,,\\
-\dfrac{1}{2}&\text{if }s=1\,,\\
0&\text{if }s\geq 3\text{ is odd}\,.
\end{cases}$$
That is, $\dfrac{x}{\exp(x)-1}=\sum\limits_{s=0}^\infty\,\dfrac{b_s}{s!}\,x^s$. Hence, (*) is equivalent to
$$\sum_{s=0}^\ell\,\binom{\ell}{s}\,b_s=b_\ell\text{ for }\ell=2,3,4,\ldots\,.\tag{#}$$

We shall now prove (#). From $\dfrac{x}{\exp(x)-1}=\sum\limits_{s=0}^\infty\,\dfrac{b_s}{s!}\,x^s$, we get
$$x=\big(\exp(x)-1\big)\sum\limits_{s=0}^{\infty}\,\dfrac{b_s}{s!}\,x^s=\left(\sum_{r=1}^{\infty}\,\frac{x^r}{r!}\right)\,\left(\sum\limits_{s=0} ^{\infty}\,\dfrac{b_s}{s!}\,x^s\right)\,.$$
This yields
$$x=\sum_{\ell=0}^\infty\,\frac{x^\ell}{\ell!}\,\sum_{s=0}^{\ell-1}\,\binom{\ell}{s}\,b_s\,.$$
This shows that
$$\sum_{s=0}^{\ell-1}\,\binom{\ell}{s}\,b_s=0\text{ for }\ell=2,3,4,\ldots\,.$$
This is equivalent to (#).

Therefore, using ($\star$), we see that $S_n^0=n$ and for an integer $p\geq 1$, we have
$$S_n^p=\frac{n^{p+1}}{p+1}+\frac{n^p}{2}-\sum_{\ell=1}^{\left\lfloor\frac{p}{2}\right\rfloor}\,\frac{(-1)^\ell\,p(p-1)(p-2)\cdots (p-2\ell+2)}{(2\ell)!}\,B_\ell\,n^{p+1-2\ell}\,.$$
Thus, the claim is established. Here are some examples:
$$S_n^1=\frac{n^2}{2}+\frac{n}{2}=\frac{n(n+1)}{2}\,,$$
$$S_n^2=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}=\frac{n(n+1)(2n+1)}{6}\,,$$
$$S_n^3=\frac{n^4}{4}+\frac{n^2}{2}+\frac{n}{4}=\left(\frac{n(n+1)}{2}\right)^2\,,$$
$$S_n^4=\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}\,,$$
and
$$S_n^5=\frac{n^6}{6}+\frac{n^5}{2}+\frac{5n^4}{12}-\frac{n^2}{12}=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}$$