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GENERAL

ARTICLE

Curious Consequences of Simple Sequences A K Mallik

S im p le se q u e n c e s a n d se r ie s o f n a tu r a l n u m b e rs , th e ir re c ip r o c a ls, in te g e r p o w e r s o f n a tu r a l n u m b e r s a n d th e ir r e c ip ro c a ls a r e c o n s id e r e d . S o m e in te r e stin g p h y sic a l a n d m a th e m a tic a l c o n se q u e n c e s o f th e se a r e d isc u sse d . T h e sim p lest seq u en ce is th e seq u en ce o f n a tu ra l n u m b ers 1 ;2 ;3 ;¢¢¢n . W e a ll k n ow th a t Sn

n = 1 + 2 + 3 + ::: + n = (n + 1 ) : 2

(1 )

S n ten d s to (S ee B ox 1 ). It is o b v io u s th a t S 1 = nlim !1 in ¯ n ity. N ex t let u s co n sid er th e seq u en ce o f recip ro ca ls o f n a tu ra l n u m b ers, i.e., 1 , 1 / 2 , 1 / 3 , ...1 / n . It h a s n o t b een p o ssib le to ex p ress th e su m H

n

= 1 + 1 = 2 + 1 = 3 + ::: + 1 = n

A K Mallik is a professor at IIT, Kanpur. His research interests are mechanical vibration, vibration control, nonlinear vibration and chaos in mechanical systems, kinematics and dynamics of machinery.

(2 )

in clo sed fo rm a s in eq u a tio n (1 ). T h e in ¯ n ite series H n ;n ! 1 , is ca lled th e h a rm o n ic series. S in ce in series (2 ), ea ch term is sm a ller th a n its p red ecesso r, w e m ay H n ex ist?" T h e a n sw er is \ N o " . a sk \ D o es H 1 = nlim !1 N ich o la s O resm e (1 3 2 3 { 1 3 8 2 ), a m u lti-fa ceted g en iu s, p rov ed it a s fo llow s: H

1

= 1+

1 1 1 1 1 1 1 + + + + + + + ¢¢¢ 2 3 4 5 6 7 8 µ

¶

µ

¶

1 1 1 1 1 1 1 > 1+ + + + + + + + ¢¢¢ 2 4 4 8 8 8 8 1 1 1 > 1 + + + + ¢¢¢ 2 2 2 S o H 1 o b v io u sly ten d s to in ¯ n ity a s w e g o o n a d d in g 1 / 2 in d e¯ n itely (see B ox 2 ).

RESONANCE  January 2007

Keywords Simple sequences, natural numbers, primes, Euler– Mascheroni constant.

23

GENERAL ARTICLE

B ox 1. Gauss virtually obtained this result when he was 7 years old. All the students of his class were asked to write the numbers 1 to 100 and add them up. Gauss quickly ¯gured out 1 + 99 = 100, 2 + 98 = 100, 3 + 97 = 100, 49 + 51 = 100. That leaves a 100 at the end and a 50 in the middle. Thus the sum is ¯fty 100's (or hundred 50's) and one ¯fty. In other words, the sum is one hundred one ¯fty's (n = 100 in equation (1) ) = 5,050.

C o n se q u e n c e o f D iv e r g e n c e o f H

1

L et u s co n sid er th e p o ssib le ov erh a n g o f a sta ck o f p lay in g ca rd s k ep t o n a h o rizo n ta l ta b le. A ssu m in g th e ca rd s to b e u n ifo rm a n d o f len g th L , th e m a x im u m p o ssib le ov erh a n g o f a sin g le ca rd L 1 is o b v io u sly L = 2 , w h en th e cen tre o f g rav ity o f th e ca rd co in cid es w ith th e ed g e o f th e ta b le. W ith tw o ca rd s (see F igu re 1 ), w h en th e co m b in ed cen tre o f g rav ity a g a in co in cid es w ith th e ed g e o f th e ta b le, th e to ta l len g th o f ov erh a n g is L2 =

L 2

µ

1+

¶

1 : 2

B ox 2. Another Proof: H

1

=1+

1 1 1 + + + ::: 2 3 4

Multiplying and dividing each term by 2 = = =

o r; T h u s, H

26

O 1

H

O 1

>

1 + 3 1 1 + + 2 4 H 1E : 1+

1 + ¢¢¢ 5 1 + ¢¢¢ 6

a lso ten d s to in ¯ n ity.


GENERAL

C o n se q u e n c e o f D iv e r g e n c e o f H

ARTICLE

O 1

C ro ssin g a D e se r t: L et u s n ow co n sid er th e fo llow in g p ro b lem . S u p p o se w e h av e a n u m b er o f jeep s, ea ch o f w h ich w ith its fu el ta n k fu ll ca n co v er a d ista n ce L . In th is d iscu ssio n , w e sh a ll refer to th e v o lu m e o f fu el a lso in term s o f L . W ith N n u m b er o f jeep s, w h a t is th e m a x im u m len g th o f a d esert th a t ca n b e cov ered b y o n e jeep w ithou t an y jeep gettin g stran ded in the desert? O f co u rse, th e jeep s a re a llo w ed ex ch a n g e o f fu el w ith in th e d esert. W ith o n e jeep , o b v io u sly o n e ca n cro ss a d esert o f len g th L 1 = L . W ith tw o jeep s, ¯ rst b o th jeep s trav el a d ista n ce L = 3 . A t th is p o in t, th e seco n d jeep tra n sfers L = 3 fu el to th e ¯ rst jeep a n d retu rn s (h a s ju st su ± cien t fu el to d o th a t). T h e ¯ rst jeep is n ow fu ll a n d ca n trav el a fu rth er d ista n ce L . S o th e to ta l d³ista n ce ´ th a t ca n b e 1 cov ered b y th e ¯ rst jeep is L 2 = L 1 + 3 . W ith 3 jeep s, a ll th e jeep s tra v el u p to L = 5 . A t th is p o in t, jeep 3 tra n sfers L = 5 fu el to jeep s 1 a n d 2 ea ch , w h ich b eco m e fu ll. J eep s 1 a n d 2 trav el a fu rth er d ista n ce L = 3 w h en jeep 2 tra n sfers L = 3 fu el to jeep 1 , w h ich a g a in b eco m es fu ll.. A t th is p o in t, jeep 2 retu rn s em p ty to th e lo ca tio n o f jeep 3 . J eep 1 n ow h a s a fu ll ta n k a n d ca n trav el a fu rth³er d ista n ce´ L . S o jeep 1 trav els a to ta l d ista n ce L 3 = L 1 + 13 + 15 , w ith jeep s 2 a n d 3 retu rn in g to th e b a se b y sh a rin g eq u a lly th e fu el rem a in in g in jeep 3 . T h u s, w ith n jeep s (a n d , o f co u rse, w ith en o u g h fu el), o n e jeep ca n cro ss a d esert o f len g th µ

L

n

= L

1+

1 1 1 + + :::: + 3 5 2n ¡ 1

¶

= LH

O n

:

S in ce H nO ten d s to in ¯ n ity w ith in crea sin g n , o n e ca n d ev ise a stra teg y to cro ss a d esert o f a n y len g th w ith o u t a n y jeep g ettin g stra n d ed in th e d esert.


27


Table 1.

n

Hn

ln(n)

Hn - ln(n)

100

5.187…

4.606…

0.581

1,000

7.486…

6.909…

0.575

1,000,000 14.392… 13.818… 0.574

R a te o f D iv e r g e n c e o f H

n

T h e su m H n d iv erg es v ery slow ly.w ith in crea sin g n . T able 1 co n v in ces u s th a t H n d iv erg es a s ln (n ) : In 1 9 6 8 , it w a s sh o w n th a t fo r H n to ex ceed 1 0 0 , n m u st b e a 4 4 -d ig it n u m b er! If th is resu lt is to b e o b ta in ed b y b ru te fo rce co m p u ta tio n b y a d d in g a term a n d ch eck in g , th en (a ssu m in g a d d itio n o f o n e term ta k es 1 0 ¡9 sec b y a g o o d co m p u ter) th e a d d itio n o f 1 0 4 4 n u m b ers w ill ta k e 1 0 3 5 sec, w h ich is m o re th a n 1 0 2 9 y ea rs (m u ch m o re th a n th e estim a ted a g e o f th e u n iv erse)! T h o u g h H n cro sses a ll in teg ers a s n in crea ses, it n ev er eq u a ls a n y in teg er. In fa ct, it is n o t d i± cu lt to sh ow th a t th e su m o f th e recip ro ca ls o f a n y n u m b er o f co n secu tiv e in teg ers ca n n o t b e a n in teg er, a s th e n u m era to r w ill b e o d d a n d th e d en o m in a to r w ill b e ev en . C o n se q u e n c e o f S lo w D iv e r g e n c e o f H

n

(i) R eco rd R a in fa ll L et u s co n sid er th e d a ta o f a n n u a l ra in fa ll a t a p la ce. W e d e¯ n e a y ea r a s a `reco rd ' y ea r, if th e ra in fa ll in th a t y ea r is g rea ter th a n in a ll p rev io u s y ea rs. T h e ex p ected n u m b er o f su ch reco rd y ea rs in n y ea rs is H n . T h e ¯ rst y ea r, b y d e¯ n itio n , is a reco rd y ea r. In th e seco n d y ea r, th e ra in fa ll (a ssu m in g d i® eren t a m o u n t o f ra in fa ll in ev ery y ea r) m ay b e m o re o r less th a n th a t in th e ¯ rst y ea r. S o th e ex p ected n u m b er o f reco rd y ea rs a t th e en d o f th e seco n d y ea r is 1 + 12 . C o n sid erin g th e ¯ rst th ree y ea rs,

28


GENERAL

ARTICLE

th e p ro b a b ility o f th e th ird y ea r h av in g th e m a x im u m ra in fa ll is 13 . T h is ca n b e ea sily seen b y co n sid erin g d ifferen t a rra n g em en ts o f th ree d i® eren t a m o u n ts o f ra in fa ll in th ree y ea rs. T h u s, th e ex p ected n u m b er o f reco rd y ea rs a t th e en d o f th e th ird y ea r is 1 + 12 + 13 . E x ten d in g th is a rg u m en t fo r n y ea rs, w e g et th e ex p ected n u m b er o f reco rd y ea rs in n y ea rs to b e H n . T h e d a ta fo r 1 6 0 y ea rs o f ra in fa ll in th e C en tra l P a rk o f N ew Y o rk C ity sh ow s 5 reco rd y ea rs w h erea s H 1 6 0 = 5 :6 5 : H ow ev er, th e d a ta fo r 2 34 y ea rs o f ra in fa ll a t O x fo rd sh ow s o n ly 5 reco rd y ea rs (H 2 3 4 = 6 :0 3 ), p ro b a b ly co n ¯ rm in g th e a g e-o ld say in g \ E n g lish w ea th er is m o st u n p red icta b le" . W e m ig h t in fer th a t u n less th e clim a te is ta m p ered w ith , th ere sh o u ld n o t b e freq u en t a p p ea ra n ces o f reco rd ra in fa ll. T h e arg u m en t g iv en a b ov e, o b v io u sly, ca n n o t b e a p p lied to a th letic reco rd s, w h ich a re d icta ted b y so m a n y o th er fa cto rs (w h ich a re n o t u n b ia sed ), lik e im p rov ed tech n iq u e, sp o n so r's in cen tiv e, etc. (ii) D e s tr u c tiv e T e stin g S u p p o se w e h av e a sa m p le o f n n o m in a lly id en tica l stru ctu ra l elem en ts, a n d th e low est stren g th o f th e lo t h a s to b e d eterm in ed b y d estru ctiv e testin g . T h e test is co n d u cted a s fo llow s. T h e ¯ rst sa m p le is b ro k en a n d th e lim itin g lo a d is d eterm in ed a s L 1 . T h e seco n d sp ecim en is lo a d ed u p to L 1 . If it d o es n o t b rea k , th en it is set a sid e. If it b rea k s a t a low er va lu e, say L 2 , th en th e th ird sp ecim en is lo a d ed u p to o n ly L 2 . If th e sa m p les ca n b e ta k en a s a ra n d o m seq u en ce, th en b y fo llow in g th is d estru ctiv e testin g p ro ced u re, o n e ex p ects to b rea k o n ly H n sp ecim en s w h ile testin g n sp ecim en s. T h e lo g ic is ex a ctly th e sa m e a s d iscu ssed in th e ra in fa ll reco rd . R eferrin g to T able 1 , o n e ex p ects to b rea k a b o u t 7 o r 8 m em b ers to d eterm in e th e m in im u m stren g th fo r a sa m p le o f 1 0 0 0 sp ecim en s.


29


R e c ip r o c a ls o f P r im e s W e k n ow th a t th e seq u en ce o f p rim e n u m b ers n ev er en d s, b u t p rim es a re sp a rsely d istrib u ted , so o n e ca n a sk \ If w e su m th e recip ro ca ls o f a ll p rim es, d o es that in ¯ n ite series co n v erg e?" T h e a n sw er is a g a in \ N o " . O b v io u sly, th is series H np d iv erg es ev en m o re slow ly th a n H n . In fa ct, th e series o f recip ro ca ls o f p rim es d iv erg es a s ln (ln (n )). J u st lik e H n , th e series H np a lso cro sses ev ery in teg er w ith in crea sin g n , b u t n ev er eq u a ls a n y in teg er. T h e p ro o f is n o t v ery d i± cu lt. O n e m ig h t p rov e th a t th e su m o f recip ro ca ls o f a n y (ev en n o n -co n secu tiv e) n u m b er o f p rim es ca n n o t b e a n in teg er. N ex t w e co n sid er a sp ecia l set o f p rim es, ca lled `tw in p rim es'. W h en tw o co n secu tiv e o d d n u m b ers a re p rim es, th en th ese a re ca lled tw in p rim es, e.g . (3 , 5 ), (5 , 7 ), (1 1 , 1 3 ),...(1 0 1 7 , 1 0 1 9 ), etc.. It is n o t y et k n ow n w h eth er tw in p rim es co n tin u e fo rev er o r n o t. B u t it is k n ow n th a t th e su m o f th e recip ro ca ls o f tw in p rim es co n v erg es: µ

H µ

+

tp

=

1 1 + 3 5

1 1 + 1019 1021

¶

µ

+

1 1 + 5 7

¶

+ :::

¶

+ :::::: = 1 :9 0 2 1 6 0 5 8 2 4 :::

T h e a b ov e n u m b er is k n ow n a s `B ru n 's co n sta n t'. If th e a b ov e su m d iv erg ed , w e co u ld h av e co n clu d ed th a t tw in p rim es d e¯ n itely co n tin u e fo rev er. N ow w e a re still u n a b le to d ecid e w h eth er tw in p rim es co n tin u e fo rev er o r n o t! (S ee B ox 3 ). B ox 3. In 1994, while computing Brun's constant and evaluating the reciprocals of twin primes (824 633 702 441, 824 633 702 443) , a °aw in the then Pentium numeric processor was detected, which was later recti¯ed by Intel. In 1995, Intel announced a pre-tax expense of US$475 million { the cost of replacement of defective chips. This is probably the largest amount of money that has ever been associated with any mathematical activity.

30


GENERAL

ARTICLE

E u le r { M a sc h e r o n i C o n s ta n t ° F ro m T able 1 w e n o te th a t th e d i® eren ce b etw een H n a n d ln (n ) is ro u g h ly eq u a l to 0 .5 7 fo r la rg e n . T o see th is clea rly, w e co n sid er F igu re 4 w h ere th e fu n ctio n y = 1 = x is p lo tted fo r 1 · x · n : N ow , a p p rox im a tin g th e a rea u n d er th e cu rv e (u p to x { a x is) b y tra p eziu m s (o n e su ch is m a rk ed b etw een x = 1 a n d 2 in th e ¯ g u re), w e ca n w rite Zn 1

dx x

µ

=

¼ ¼

¶

µ

¶

1 1 1 1 1 ln n ¼ 1+ + + + 2 µ 2 ¶ 2 2 3µ ¶ 1 1 1 1 1 1 + + ::: + + n 2 3 4 2 n ¡ 1 · µ ¶ ¸ 1 1 1 1+ 2 H n ¡ 1¡ + n n 2 1 1 H n ¡ ¡ : 2 2n

T h u s,

¼ ln n +

1 1 : + 2 2n L et ° n = H n ¡ ln n ; th is is th e d i® eren ce o f tw o d iv erg in g term s. It m ay b e sh ow n th a t ° n co n v erg es. T h e ° n is ca lled `E u ler{ M a sch ero n i C o n co n sta n t ° = nlim !1 Figure 4. sta n t'. E u ler ¯ rst ca lcu la ted it u p to ¯ v e a n d th en to H

n


Figure 4.

31


six teen d ecim a l p la ces a s ° = 0 :5 7 7 2 1 5 6 6 4 9 0 1 5 3 2 5 :::. In 1 9 9 8 , th is co n sta n t w a s co m p u ted u p to 1 0 8 £ 1 0 6 d ecim a l p la ces. B u t it is n o t y et p rov ed to b e a n irra tio n a l n u m b er. S u m s o f P o w e r s o f In te g e r s a n d R e c ip r o c a ls W e h av e a lrea d y n o ted (see eq u a tio n (1 )) th a t S n(1 ) = 1 + 2 + 3 + ::: + n =

n (n + 1 ) : 2

J a k o b B ern o u lli ca lcu la ted th e su m s o f p ow ers o f in teg ers in term s o f w h a t a re k n ow n a s B ern o u lli n u m b ers. F o r ex a m p le, if w e w rite S n(k ) = 1 k + 2 k + 3 k + ::: + n k ; th en S n(1 )

Ã Ã

n 1 = (n + 1 ) = 2 2

!

2 0

Ã

B 0n 2 +

!

2 1

!

B 1n

n (n + 1 ) (2 n + 1 ) 6 Ã Ã ! Ã ! Ã ! ! 1 3 3 3 3 2 B 0n + B 1n + B 2n = 0 1 2 3

S n(2 ) =

´ ³ ´2 n2 ³ 2 n + 2 n + 1 = S n(1 ) ; 4 w h ere th e B ern o u lii n u m b ers a re

S n(3 ) =

B B B

0 4 3

= 1; B 1 = 1= 2; B 2 = 1=6; = ¡ 1 = 3 0 ; B 6 = 1 = 4 2 ; ::::: = B 5 = B 7 = :::: = 0 :

W e m ay n o te th a t ex cep t th e ¯ rst o n e, a ll o th er o d d B ern o u lii n u m b ers a re zero . G en era lizin g th e a b ov e resu lts fo r a n y va lu e o f k is left to th e rea d er. W e h av e a lrea d y n o ted th a t th e H a rm o n ic series H 1 d iv erg es. N ow w e sh ow th a t if w e co n sid er th e su m o f th e

32


GENERAL

ARTICLE

p ow ers (m o re th a n 1 ) o f recip ro ca ls, th en it co n v erg es. T ow a rd s th is en d , w e fo llow th e p ro ced u re o f N ich o la s O resm e o u tlin ed ea rlier. L et ³ (s ) = = < = =

1 1 1 + s + s + ::: s 2 3 ¶4 µ µ ¶ 1 1 1 1 1+ + s + + ::: + s + ::: 2s 3 4s 7 2 4 1 + s + s + ::: 2 4 µ ¶ 1 2 1 1 + s¡1 + + ::: 2 2 s¡ 1 1 1 ¡ 2 s¡1 1

1+

T h u s ³ (s ) co n v erg es if

1 2 s¡

1

< 1 ; i.e., s > 1 .

E u ler w a s th e ¯ rst to ca lcu la te ³ (s ) fo r ev en va lu es o f s u p to 2 6 . H e a lso sh ow ed th a t fo r ev en va lu es o f s, o n e g ets, in term s o f B ern o u lli n u m b ers, ³ (2 n ) = (¡ 1 )

n ¡1

(2 ¼ )2 n B 2 (2 n ) !

2n

:

N o resu lt is ava ila b le fo r o d d va lu es o f s (in terestin g ly, a s m en tio n ed ea rlier, a ll o d d B ern o u lli n u m b ers a re zero fo r s m o re th a n 1 ). In 1 9 7 8 it w a s sh ow n th a t ³ (3 ) is irra tio n a l. (S ee B ox 4 ). C o n n e c tio n B e tw e e n T w o In ¯ n ite U n iv e r se s o f N a tu r a l N u m b e r s a n d P r im e s E u ler p ro d u ced th e fo llow in g rem a rka b le resu lt, sta rtin g fro m ³ (s ), co n n ectin g a ll th e n a tu ra l n u m b ers to a ll th e p rim e n u m b ers. T h is p ro cess is k n ow n a s `E ra to sth en es siev e'. T h e ¯ n a l resu lt is th e sta rtin g p o in t o f a lo t o f la ter m a th em a tics, esp ecia lly in th e th eo ry o f p rim e n u m b ers. W e sta rt w ith ³ (s ) =

1 1 1 + s + s + :::; s 1 2 3


s> 1

33


a n d th en m u ltip ly in g b o th sid es b y

1 2s

w e g et

1 1 1 1 ³ (s ) = s + s + s + ::: : s 2 2 4 6 S u b tra ctin g th is eq u a tio n from th e p reced in g o n e, w e o b ta in µ

1¡

1 2s

¶

³ (s ) =

1 1 1 1 1 + + + + + ::: : 1s 3s 5s 7s 9s

B ox 4. Euler in 1735 (using a non-rigorous method) ¯rst obtained the value of ³ (2) , which enhanced his reputation enormously as all great mathematicians before him failed at this problem. He started from the in¯nite series sin (x ) = x ¡

x3 x5 x7 + ¡ + ::: : 3! 5! 7!

Dividing both sides by x x2 x4 x6 sin x =1¡ + ¡ + ::: : x 6 120 5040 Substituting y = x 2 ,

1¡

y y2 y3 sin x : + ¡ + ::: = x 6 120 5040

Now the right hand side is zero for x = n ¼ ; n = 1; 2; 3;:::: , i.e. with y = n 2 ¼ 2 . So Euler treated the in ¯ n ite polynomial equation (with right hand side zero) as having roots y n = n 2 ¼ 2 : Then he used the result valid for a ¯nite polynomial equation, which says that the sum of the reciprocal of the roots is governed by only the coe±cient of y (with the leading term as unity) . Thus, X n

1 1 = ; yn 6

X or;

³ (2) = n

¼2 1 = ! n2 6

If we are surprised to see the appearance of ¼ (the ratio of circumference to diameter of a circle) in this summation, then it will be quite astonishing to know that 6= ¼ 2 is also the probability of two positive integers taken at random to be coprime, i. e. , not to have any common factor! Since the sum of the reciprocals of primes did not converge, Euler concluded that the number of primes is more than the number of perfect squares (of course, both are in¯nite) .

34


GENERAL

M u ltip ly in g b o th sid es b y µ

1 1 1¡ s s 3 2

1 3s

¶

³ (s ) =

ARTICLE

w e g et

1 1 1 + s + + ::: : s 3 9 15s

A g a in su b tra ctin g th e la st eq u a tio n fro m th e p reced in g o n e, w e o b ta in µ

1¡

1 2s

¶µ

1¡

¶

1 1 1 1 1 ³ (s ) = s + s + s + + ::: : s 3 1 5 7 11s

P ro ceed in g in th is m a n n er, a ll th e p rim e n u m b ers a p p ea r o n th e left h a n d sid e, a n d th e rig h t h a n d sid e b eco m es 1 . T h u s, ¯ n a lly Ã

1 1¡ s p

¦p or

X n

!

³ (s ) = 1

1 1 ³ = s n ¦p 1 ¡

1 ps

´ :

T h e left h a n d sid e is su m m ed ov er a ll th e p o sitiv e in teg ers w h erea s, o n th e rig h t h a n d sid e, w e ta k e th e p ro d u ct ov er a ll p rim es. A lte r n a tin g S e r ie s L et u s co n sid er th e fo llow in g in ¯ n ite g eo m etric series 1+

1 1 1 1 + + + ::: = 2 4 8 1¡

1 2

= 2:

S in ce th is series co n v erg es w ith a ll th e term s h av in g p o sitiv e sig n , if th e a ltern a te term s a re w ritten w ith o p p o site sig n s, th en a lso it w ill co n v erge, i.e., 1¡

1 1 1 1 ¡ ¡ ::: = + 2 4 8 1+


1 2

=

2 : 3

35


S u ch a series is ca lled a b so lu tely co n v erg en t, a n d it a lw ay s co n v erg es to th e sa m e va lu e in d ep en d en t o f o rd erin g o f th e term s. W e h av e a lrea d y seen th a t th e h a rm o n ic series d iv erg es. B u t if th e sig n in fro n t o f th e a ltern a te term s a re m a d e o p p o site th en th e a ltern a tin g h a rm o n ic series

H

A 1

= 1¡

1 1 1 1 1 ¡ ¡ + + + ::: 2 3 4 5 6

w h ich is th e d i® eren ce o f tw o d iv erg in g series, v iz. H 1O a n d H 1E , co n v erg es a n d is ca lled a co n d itio n a lly co n v erg en t series. A co n d itio n a lly co n v erg en t series co n v erg es to a va lu e w h ich d ep en d s o n th e o rd erin g o f th e term s. F o r ex a m p le, if co m p u ted a s w ritten a b o v e, H 1A co n v erg es to ln 2 . T o sh ow th is w e sta rt w ith ln (1 + x ) = 1 ¡

x2 x3 x4 ¡ + + ::: 2 3 4

¡ 1 < x · 1

O n su b stitu tin g x = 1 , w e g et th e d esired resu lt. H ow ev er, if w e rew rite th e sa m e series w ith a d i® eren t o rd erin g o f term s a s H

A 1

µ

=

1 1¡ 2

¶

1 ¡ + 4

µ

1 1 ¡ 3 6

¶

1 ¡ + 8

µ

1 1 ¡ 5 10

¶

¡

1 + ::: 12 th en w e g et th is to b e = = =

36

1 2 1 2 1 2

¡

1 1 1 1 1 ¡ ¡ + + + ::: 6 8 10 12 µ 4 ¶ 1 1 1 1 1 ¡ ¡ 1¡ + + + ::: 2 3 4 5 6 ln 2 :


GENERAL

ARTICLE

S u rp risin g ly, o n e ca n m a k e H 1A co n v erg e to ln 3 b y th e fo llow in g o rd erin g o f th e term s ln 3 = 1 + µ

1 1 1 1 1 1 1 1 1 1 1 ¡ ¡ ¡ ¡ + + + + + + + 3 2 5 7 4 9 11 6 13 15 8 ¶ 1 1 + 17 19 µ 1 1 1 1 1 1 1 1 ¡ ¡ ¡ + + + + + + 21 10 23 25 12 27 29 14 ¶ 1 1 1 1 1 ¡ + + + + ::: 31 33 16 35 37 N o tice th a t in a ll th e o rd erin g m en tio n ed a b ov e, th e term s w ith th e sa m e sig n a p p ea r in d ecrea sin g o rd er. S u ch a rra n g em en ts o f th e term s a re ca lled sim p le a rra n g em en ts. T h ere a re th eo rem s fo r su ch sim p le o rd erin g . In fa ct, if w e in clu d e ev en th e n o n -sim p le o rd erin g , th en th e co n d itio n a lly co n v erg en t series ca n b e m a d e to co n v erg e to a n y n u m b er! Suggested Reading [1] J Havil, GAMMA exploring Euler’s Constant, Universities Press (India) Private Limited, 2005.

Address for Correspondence A K Mallik Department of Mechanical Engineering Indian Institute of Technology Kanpur 208 016, India. Email:[email protected]

[2] J Derbyshire, Prime Obsession, Plume, Washington DC, 2004.


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