A-1-1-2-1-3-B-1-1-2-1-3-A-B-

Question Number 1997 by prakash jain last updated on 29/Oct/15

A = 1 + \frac{1}{2} + \frac{1}{3} + \dots .

B = 1 + \frac{1}{2} + \frac{1}{3} + \dots .

A - B = ?

Commented by Rasheed Soomro last updated on 30/Oct/15

I f^{'} \dots^{'} r e p r e s e n t s^{'} n o n - {e n d i n g}^{'}

A = 1 + \frac{1}{2} + \frac{1}{3} + \dots .

\dots

B = 1 + \frac{1}{2} + \frac{1}{3} + \dots .

A - B = (1 - 1) + (\frac{1}{2} - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n}) + \dots

= 0 + 0 + 0 + \dots

= 0 . \infty

= I n d e t e r m i n a t e

Answered by Yozzi last updated on 29/Oct/15

A a n d B b o t h r e p r e s e n t i n f i n i t e

h a r m o n i c s e r i e s . B o t h s e r i e s a r e t h e n

d i v e r g e n t a n d A - B i s u n d e f i n e d .

Commented by Filup last updated on 29/Oct/15

{Can}^{'} t you say :

A = B = 1 + \frac{1}{2} + \frac{1}{3} + \dots

Let S = A = B

∴ A - B = S - S = 0 ?

Commented by Filup last updated on 29/Oct/15

Similar to :

S = 1 - 1 + 1 - 1 + \dots

S_{1} = (1 - 1) + (1 - 1) + \dots = 0

S_{2} = 1 - (1 - 1) - (1 - 1) - \dots = 1

S_{3} = 0.5

∵ 1 - S = 1 - (1 - 1 + 1 - 1 + \dots)

∴ 1 - S = S

2 S = 1

S = 0, 0.5, 1

S imilar to :

A = B = 1 + \frac{1}{2} + \frac{1}{3} + \dots

(1) A - B = (1 + \frac{1}{2} + \frac{1}{3} + \dots) - (1 + \frac{1}{2} + \frac{1}{3} + \dots)

A - B = (1 - 1) + (\frac{1}{2} - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{3}) + \dots

∴ A - B = 0

o r

(2) S = A = B = 1 + \frac{1}{2} + \frac{1}{3} + \dots

∴ A - B = S - S = 0

∴ A - B = 0

Commented by Yozzi last updated on 29/Oct/15

H o w a r e w e s o c e r t a i n t h a t t h e s e r i e s

o f e a c h A a n d B h a v e t h e s a m e n u m b e r

o f t e r m s a s n \to \infty ?

E a c h s e r i e s i s o f t h e f o r m

\underset{r = 1}{\sum^{\infty}} \frac{1}{n} . T h u s, i t i s p o s s i b l e a s y o u r i g h t l y

s a i d t h a t A c o u l d e q u a l B b e a t a n

a r b i t r a r i l y l a r g e v a l u e o f n .

∴ o n e p o s s i b i l i t y i s \exists N \in N s u c h t h a t

t h e p a r t i a l s u m \underset{r = 1}{\sum^{N}} \frac{1}{n} i s b o t h e q u a l

t o A a n d B \Rightarrow A - B = 0 .

H o w e v e r, w h a t i f t h e s e r i e s e a c h

d i f f e r b y a t l e a s t o n e t e r m s o w e h a v e

N \neq M (N > M > 3) w h e r e A = \underset{r = 1}{\sum^{N}} \frac{1}{r}

a n d B = \underset{r = 1}{\sum^{M}} \frac{1}{r} . I n t h i s c a s e w e m a y h a v e f o r e x a m p l e

A = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots + \frac{1}{1234567890} + \frac{1}{1234567891} + \frac{1}{1234567892}

A = \underset{r = 1}{\sum^{1234567892}} \frac{1}{r}

a n d B = \underset{r = 1}{\sum^{1234567891}} \Rightarrow A \neq B \Rightarrow A - B \neq 0 .

I s a y A - B i s u n d e f i n e d b e c a u s e

n = \infty r e p r e s e n t s a n a r b i t r a r i l y l a r g e

n u m b e r o f t e r m s s o t h a t o n e {d o e s n}^{'} t

k n o w f o r s u r e t h a t A = B t h o u g h

A a n d B h a v e i d e n t i c a l n o t a t i o n a l

r e p r e s e n t a t i o n s .

Commented by Filup last updated on 29/Oct/15

I agree with the above . However, due to

the similariy of a pattern and the open

possible expectation that both patterns continue

either the same or similarl magnitude, I would argue

that both statements are correct such that :

F o r :

A = 1 + \frac{1}{2} + \frac{1}{3} + \dots

A = \underset{i = 1}{\sum^{n}} \frac{1}{i} = H_{n}

B = 1 + \frac{1}{2} + \frac{1}{3} + \dots

B = \underset{i = 1}{\sum^{m}} \frac{1}{i} = H_{m}

A - B = {\begin{cases} 0, n = m \\ undefined, n \neq m \end{cases}

Commented by Yozzi last updated on 29/Oct/15

C o r r e c t .

Commented by Rasheed Soomro last updated on 29/Oct/15

O n t h e c o m m e n t o f Y o z i

Y o u h a v e o b j e c t e d a b o u t t o b e b o t h s e r i e s

o f s a m e n u m b e r o f t e r m s . I a g r e e t h a t o n e i n f i n i t y m a y n o t

b e e q u a l t o o t h e r i n f i n i t y h o w e v e r i n t h i s c a s e I t h i n k b o t h

i n f i n i t i e s a r e t h e s a m e a s o f t h e i n f i n i t y o f N b e c a u s e

t h e y a r e c o n c e r n e d w i t h c o u n t i n g . (O r a l l i n f i n i t i e s

a r e c o n c e r n e d w i t h c o u n t i n g ? ? ?) T h i s c a n b e s a i d i n b e t t e r

w a y t h a t t e r m s o f a n y o f t h e s e r i e s c a n b e m a t c h e d w i t h

e l e m e n t s o f N .

{C o u d n}^{'} t w e s a y t h a t w e a r e c e r t a i n t h a t b o t h s e r i e s h a v e

n o e n d ? w h e n w e c o n s i d e r t h a t b o t h s e r i e s m a y h a v e

d i f f e r e n t n u m b e r o f t e r m s w e a r e r e a l l y t a l k i n g i n \underline{f i n i t e s e n s e} .

Commented by Yozzi last updated on 29/Oct/15

{Y o u}^{'} r e c o r r e c t, b u t t h e n a g a i n i f w e

s a y A = B c a n y o u p r o v e t o m e t h a t

r e g a r d l e s s o f h o w l a r g e t h e t e r m s

o f e a c h o f t h e s e r i e s g e t s, A a l w a y s

e q u a l s B ? T h e n o t a t i o n o n l y s u g g e s t s

h o w e a c h s e r i e s i s g e n e r a t e d . T h e

n o t a t i o n d o e s n o t e x p l i c i t l y i n d i c a t e

w h e t h e r o r n o t t h e t w o s e r i e s c o n v e r g e

t o e a c h o t h e r . B u t, {t h e r e}^{'} s a c h a n c e t h a t

I a m m i s i n t e r p r e t a t i n g (p r o b a b l y) h o w t h e i d e a

o f i n f i n i t y p l a y s o u t i n d e c i d i n g i f

A = B . M y u n d e r s t a n d i n g o f p r o v i n g

A = B i s t o s h o w t h a t t h e i n f i n i t y c o n c e r n e d

w i t h A e q u a l s t h a t c o n c e r n e d w i t h B .

Commented by Rasheed Soomro last updated on 30/Oct/15

P l s e e m y c o m m e n t f o l l o w i n g t h e Q u e s t i o n .

I h a v e s u p p o r t e d y o u e v e n i n c a s e b o t h i n f i n i t i e s

r e g a r d i n d A a n d B a r e s a m e a s o f N .

Commented by 123456 last updated on 31/Oct/15

also if a series is conditionaly convergence

you can change it value by reorganizing it

a serie is conditionaly convergence if

Σ a_{n} converge but Σ ∣ a_{n} ∣ not

if Σ ∣ a_{n} ∣ converge the covergence are called absolute covergenceu

Answered by Rasheed Soomro last updated on 01/Nov/15

A = 1 + \frac{1}{2} + \frac{1}{3} + \dots .

\dots .

B = 1 + \frac{1}{2} + \frac{1}{3} + \dots .

A - B = (1 - 1) + (\frac{1}{2} - \frac{1}{2}) + \dots (\frac{1}{n} - \frac{1}{n}) + \dots

= 0 + 0 + 0 + \dots

= 0 . \infty

= i n d e t e r m i n a t e