查看“︁肯普納級數”︁的源代码

'''肯普納級數'''（{{lang-en|Kempner series}}）是[[十進制]]寫法不含數字9的正整數的[[倒數]]和。用符號可寫成
:<math> \sum_{\begin{smallmatrix}n=1 \\ n \text{缺 } 9\end{smallmatrix}}^\infty \frac{1}{n} _,</math>
其中「缺9」意思是「十進制表示中，不含數字9」，下同。{{link-en|奧伯利·肯普納|Aubrey J. Kempner}}於1914年最早研究該級數。<ref name="Kempner 1914">
{{cite journal
 | last = Kempner
 | first = A. J.
 | date= 1914-02
 | title = A Curious Convergent Series
 | trans-title = 某稀奇的收斂級數
 | jstor = 2972074
 | journal = [[美國數學月刊|American Mathematical Monthly]]
 | volume = 21
 | issue = 2
 | pages = 48&ndash;50
 | publisher = Mathematical Association of America
 | location = Washington, DC
 | issn = 0002-9890
 | doi = 10.2307/2972074
 | language = en
}} </ref>肯普納級數是由[[調和級數]]刪走含數字9的項所得，但肯普納級數[[收斂]]，調和級數則[[發散]]。肯普納證明，級數之和小於90。羅伯特·貝利<ref name = "Baillie 1979"/>證明，級數準確到小數點後20位的值為{{gaps|22.92067|66192|64150|34816}}{{OEIS|A082838}}。

直觀理解，級數收斂是因為大部分「大數」都有齊0至9的全部數字。例如，[[離散型均勻分佈|均勻隨機]]選一個100位的正整數，很易包含至少一個數字9，於是級數不計該數的倒數。

施梅爾策與貝利<ref name="Schmelzer and Baillie 2008">
{{cite journal
 | last1 =Schmelzer
 | first1 =Thomas
 | first2=Robert 
 | last2=Baillie
 | date= 2008-06
 | title = Summing a Curious, Slowly Convergent Series
 | trans-title = 求某稀奇而收斂得慢的級數和
 | journal = [[美國數學月刊|American Mathematical Monthly]]
 | volume =115
 | issue = 6
 | pages = 525&ndash;540
 | publisher = Mathematical Association of America
 | location = Washington, DC
 | issn = 0002-9890
 | mr = 2416253
 | jstor = 27642532
 | language = en
}}</ref>找到高效[[算法]]，給定任意數字串為輸入，計算缺該串的正整數倒數和。此問題推廣了原本的級數求值問題。舉例，考慮所有缺數字串「42」的正整數<math>n</math>，其倒數和約為{{gaps|228.44630|41592|30813|25415}}。又舉例，缺數字串「314159」的正整數倒數和約為{{gaps|2302582.33386|37826|07892|02376}}。（上述數值皆四捨五入至末位。）

== 命名 ==

許多文獻中，級數未有命名。<ref>{{cite journal |journal = 南京工学院学报 |issue = 3 |year = 1985 |url = http://journal.seu.edu.cn/oa/darticle.aspx?type=view&id=198503012 |title = 调和级数的收敛子级数的和 |author1 = 赵显曾 |author2 = 冯世 |author3 = 程乃毅 |access-date = 2021-11-08 |archive-date = 2021-11-08 |archive-url = https://web.archive.org/web/20211108182653/http://journal.seu.edu.cn/oa/darticle.aspx?type=view&id=198503012 |dead-url = no }}</ref>[[MathWorld]]用{{lang|en|Kempner series}}為條目名。<ref>{{cite mathworld|title=Kempner series|urlname=KempnerSeries}}</ref>朱利安·哈維爾所著《伽瑪》（論[[歐拉-馬斯刻若尼常數]]）亦採用同一名稱。<ref>{{cite book | last = Havil | first = Julian | title = Gamma: Exploring Euler's Constant | url = https://archive.org/details/gammaexploringeu0000havi | trans-title = 伽瑪：探索歐拉常數| publisher = Princeton University Press | location = Princeton | year = 2003 | isbn = 978-0-691-09983-5 |language = en}}</ref>{{Rp|31–33}}

== 收斂 ==

肯普納對數列收斂之證明<ref name="Kempner 1914" />，載於若干教科書，如[[戈弗雷·哈羅德·哈代|哈代]]與[[愛德華·梅特蘭·賴特|賴特]]合著《數論導論》<ref>{{cite book | author1-last = Hardy | author1-first = G. H. |author1-link = 戈弗雷·哈羅德·哈代| author2-first = E. M. |author2-last = Wright|author2-link = 愛德華·梅特蘭·賴特 | title = An Introduction to the Theory of Numbers |trans-title = 數論導論| publisher = Clarendon Press | location = Oxford | year = 1979 | edition = 5th | isbn = 0-19-853171-0 | url = https://archive.org/details/introductiontoth00hard |language = en}}</ref>{{Rp|120}}，亦是[[湯姆·麥克·阿波斯托|阿波斯托]]《數學分析》的習題<ref>{{cite book | last = Apostol | first = Tom |author-link = 湯姆·麥克·阿波斯托| title = Mathematical Analysis | url = https://archive.org/details/mathematicalanal02edtomm |trans-title = 數學分析| publisher = Addison&ndash;Wesley | location = Boston | year = 1974 | isbn = 0-201-00288-4 |language = en}} 該書有中譯本：<br>{{cite book|title = 数学分析| publisher = 机械工业出版社 | author1 = Tom M. Apostol（著）| author2 = 邢富冲 |author3 = 邢辰 |author4 = 李松洁 |author5 = 贾婉丽（译）|isbn = 7-111-18014-3 | series = 华章数学译丛 |year = 2006}}</ref>{{Rp|212}}。證明如下。

將級數各項按分母的位數分組。由[[乘法原理]]，缺「9」的<math>n</math>位正整數共有<math>8 \times 9^{n-1}</math>個，因為最高位有8個選擇（1至8，首位不為零），而其後<math>n - 1</math>位，每位有9種選擇（0至8），且各位的選擇互相獨立。任何<math>n</math>位數皆不小於<math>10^{n-1}</math>，故其倒數至多為<math>10^{1-n}</math>。所以，缺「9」的<math>n</math>位正整數之倒數，對級數的貢獻，至多是<math>8 \times \left(\frac{9}{10}\right)^{n-1}</math>。因此，將各組貢獻加總，全個級數至多為

:<math>8 \sum_{n=1}^\infty \left(\frac{9}{10}\right)^{n-1} = 80.</math>

若將禁止出現的「9」換成其他非零數字，則同樣的論證仍成立。至於缺「0」的情況，缺「0」的<math>n</math>位正整數共有<math>9^n</math>個，故缺「0」正整數的倒數和至多為：

:<math>9 \sum_{n=1}^\infty \left(\frac{9}{10}\right)^{n-1} = 90.</math>

若刪去含有某串<math>k</math>位[[字串|子字串]]的項，例如忽略所有分母含子字串「42」的項，則級數同樣收斂。證明方法幾乎一樣<ref name="Schmelzer and Baillie 2008" />，先觀察在<math>10^k</math>進制中，刪去含有該字串為「位」的項，則前述證明適用，證出新級數收斂。但是，新級數比欲證收斂的級數更大，原因是欲證收斂的級數中，不僅刪走以該字串為<math>10^k</math>進制位的項，還刪走了跨<math>10^k</math>進制位而含該字串的項。接續前一個例子，百進制的新級數略過4217（百進制的首位是42）和1742（百進制的末位是42），但未略過1427，而欲證收斂的級數中，連1427也一併略去。

巴基爾·法喜<ref>
{{cite journal
 | last = Farhi
 | first = Bakir
 |date=December 2008
 | title = A Curious Result Related to Kempner's Series 
 | url = https://archive.org/details/sim_american-mathematical-monthly_2008-12_115_10/page/933
 | trans-title = 有關肯普納級數的稀奇結果
 | journal = American Mathematical Monthly
 | volume = 115
 | issue = 10
 | pages = 933&ndash;938
 | publisher = Mathematical Association of America
 | location = Washington, DC
 | issn = 0002-9890
 | mr = 2468554
 | bibcode = 2008arXiv0807.3518F
 | jstor = 27642640
 | arxiv=0807.3518
 | language = en
}}</ref>研究恰有<math>n</math>個數字<math>d</math>（滿足<math>0\le d \le 9</math>）的正整數倒數和<math>S(d, n)</math>，此為肯普納級數的推廣，因為原級數即為<math>S(9,0)</math>。法喜證明，對每個<math>d</math>，數列<math>S(d, n)</math>由<math>n  = 1</math>起取值遞減，且當<math>n</math>趨向無窮大時，收斂到<math>10 \ln 10</math>。不過，數列一般並非由<math>n = 0</math>起遞減，例如原級數值為<math>S(9,0) \approx 22.921 < 23.026 \approx 10 \ln 10</math>，比<math>n\ge 1</math>時任意一個<math>S(9, n)</math>更小。

== 數值方法 ==
級數收斂得很慢。貝利<ref name="Baillie 1979" />寫道，即使計算前10<sup>24</sup>項和，其後餘項仍超過1。<ref name = "Bai79Errata">
{{cite journal
 | date=December 1980
 | title = ERRATA
 | url=https://archive.org/details/sim_american-mathematical-monthly_1980-12_87_10/page/n89
 | trans-title = 勘誤
 | journal = American Mathematical Monthly
 | volume = 87
 | issue = 10
 | page = 866
 | publisher = Mathematical Association of America
 | location = Washington, DC
 | issn = 0002-9890
 | doi=10.2307/2320815
 | language = en 
}}</ref>

80是很粗略的上界。弗蘭克·厄文較仔細地分析後<ref>
{{cite journal
 | last = Irwin
 | first = Frank
 |date=May 1916
 | title = A Curious Convergent Series
 | trans-title = 某稀奇的收斂級數
 | jstor = 2974352
 | journal = American Mathematical Monthly
 | volume = 23
 | issue = 5
 | pages = 149–152
 | publisher = Mathematical Association of America
 | location = Washington, DC
 | issn = 0002-9890
 | doi = 10.2307/2974352
 | language = en
}}</ref>，證明級數值接近23。此後，再由貝利改進到前述的22.92067…。<ref name="Baillie 1979" />

貝利{{refn|name="Baillie 1979"|{{cite journal
 | last =Baillie
 | first =Robert
 | date= 1979-05
 | title = Sums of Reciprocals of Integers Missing a Given Digit
 | url =https://archive.org/details/sim_american-mathematical-monthly_1979-05_86_5/page/372
 | trans-title = 缺給定數字的整數倒數和
 | jstor =2321096
 | journal = American Mathematical Monthly
 | volume = 86
 | issue = 5
 | pages = 372–374
 | publisher = Mathematical Association of America
 | location = Washington, DC
 | issn = 0002-9890
 | doi =10.2307/2321096
 | language = en
}}經<ref name = "Bai79Errata"/>訂正。}}以<math>s(i, j)</math>表示缺某指定數字的所有<math>i</math>位正整數的<math>j</math>次方倒數和，然後推導出，只要有齊<math>s(i, j+n)</math>對所有非負整數<math>n</math>的值，就能[[遞歸關係|遞歸計算]]<math>s(i+ 1, j)</math>。於是，祗需較少的計算，已得到原級數<math>s(1, 1) + s(1, 2)+ \cdots </math>的準確估計。

== 參見 ==
* [[倒數和收斂]]
* {{le|倒數和列表|List of sums of reciprocals}}

== 參考文獻 ==
{{reflist}}

[[Category:级数]]
[[Category:数值分析]]