一个与多项式分拆有关的级数题 - Eufisky - The lost book
西西爆难积分题求解
一个很火的积分题

一个与多项式分拆有关的级数题

Eufisky posted @ 2014年8月29日 17:54 in 数学分析 with tags 级数 分拆 , 966 阅读
求证:对于$\forall k\in N_+$,必有\[\sum\limits_{n = 1}^\infty  {\frac{{{n^k}}}{{n!}}}\]是$e$的整数倍.
证明.先证明一个引理:对$\forall k\in N_+$,均有${n^k} = {a_0}n + {a_1}n\left( {n - 1} \right) + {a_2}n\left( {n - 1} \right)\left( {n - 2} \right) +  \cdots  + {a_k}n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k+1} \right)$成立,其中$a_1+a_2+\ldots+a_i+\ldots+a_{k-1}(i=0,1,2,\ldots,k)$为整数(事实上,$a_i$均为正整数,$i\in N^+$).
证明.写得规范点,我们有:\[{x^k} = {a_0}x + {a_1}x\left( {x - 1} \right) + {a_2}x\left( {x - 1} \right)\left( {x - 2} \right) +  \cdots  + {a_{k-1}}x\left( {x - 1} \right)\left( {x - 2} \right) \cdots \left( {x - k+1} \right).\]
比较两边$x^k$的系数得知$a_{k-1}=1$,即
\begin{align*}&x\left( {x - 1} \right)\left( {x - 2} \right) \cdots \left( {x - k + 1} \right) = {x^k} - {a_{k - 2}}x\left( {x - 1} \right)\left( {x - 2} \right) \cdots \left( {x - k + 2} \right) \\&-  \cdots  - {a_2}x\left( {x - 1} \right)\left( {x - 2} \right) - {a_1}x\left( {x - 1} \right) - {a_0}x.\end{align*}
 
再比较两边$x^{k-1}$的系数得$a_{k-2}=\frac{k(k-1)}{2}$,由此
\[\frac{{{{\left( {1 + 2 +  \cdots  + \left( {k - 1} \right)} \right)}^2} - \left( {{1^2} + {2^2} +  \cdots  + {{\left( {k - 1} \right)}^2}} \right)}}{2} = \left( {1 + 2 +  \cdots  + \left( {k - 2} \right)} \right){a_{k - 2}} - {a_{k - 3}}.\]故\[{a_{k - 3}} = \frac{{\left( {3k - 5} \right)\left( {k - 2} \right)\left( {k - 1} \right)k}}{{24}}.\]
一般地,考察等式两边$x^n$的系数我们有
\begin{align*}&{\left( { - 1} \right)^{k - n}}\sum\limits_{1 \le {k_1} < {k_2} <  \cdots  < {k_{k - n}} \le k - 1} {{k_1}{k_2} \cdots {k_{k - n}}}  = {\left( { - 1} \right)^{k - n}}{a_{k - 2}}\sum\limits_{1 \le {k_1} < {k_2} <  \cdots  < {k_{k - n - 1}} \le k - 2} {{k_1}{k_2} \cdots {k_{k - n - 1}}} \\&+ {\left( { - 1} \right)^{k - n - 1}}{a_{k - 3}}\sum\limits_{1 \le {k_1} < {k_2} <  \cdots  < {k_{k - n - 2}} \le k - 3} {{k_1}{k_2} \cdots {k_{k - n - 2}}}  +  \cdots  + {\left( { - 1} \right)^2}{a_n}\sum\limits_{1 \le {k_1} \le n} {{k_1}} {x^n} + {\left( { - 1} \right)^1}{a_{n - 1}}.\end{align*}
依此递推式得$a_0=1.$且$a_i,i=0,1,\ldots,k-2$均为正整数.
回到原题,我们有
\begin{align*}\sum\limits_{n = 1}^\infty  {\frac{{{n^k}}}{{n!}}}  &= \sum\limits_{n = 1}^\infty  {\frac{{{a_0}n + {a_1}n\left( {n - 1} \right) + {a_2}n\left( {n - 1} \right)\left( {n - 2} \right) +  \cdots  + {a_{k - 1}}n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k + 1} \right)}}{{n!}}} \\&= {a_0}\sum\limits_{n = 1}^\infty  {\frac{1}{{\left( {n - 1} \right)!}}}  + {a_1}\sum\limits_{n = 2}^\infty  {\frac{1}{{\left( {n - 2} \right)!}}}  + {a_2}\sum\limits_{n = 3}^\infty  {\frac{1}{{\left( {n - 3} \right)!}}}  +  \cdots  + {a_{k - 1}}\sum\limits_{n = k}^\infty  {\frac{1}{{\left( {n - k} \right)!}}} \\&= \left( {{a_0} + {a_1} + {a_2} +  \cdots {a_{k - 1}}} \right)e = Ne.\end{align*}
Avatar_small
UPMSP Plus One Previ 说:
2022年8月15日 23:15

For all of the students enrolled in these institutions, this board is drafting a syllabus from Class I to Class XII. There are several schools in the state of Uttar Pradesh that are associated with this board. Every year, this board also holds exams for the ninth and eleventh grades, in which a large number of students participate. These exams' UPMSP Plus One Question Paper 2023 was released in June. UPMSP Plus One Previous Paper 2023 As we know, the UPMSP Plus One Important Question Paper 2023 for class 11th of the UP Board was announced in the month of June the previous year, and we presume, there were 26 lakh students who appeared in the Class 11th UP board examinations, and the pass % of the entire students.


登录 *


loading captcha image...
(输入验证码)
or Ctrl+Enter