P-n-P-n-Q-n-P-n-P-2-3-5-7-Q-1-4-6-8-Is-P-gt-Q-Is-Q-gt-P-

Question Number 15497 by FilupS last updated on 11/Jun/17

P = \underset{n \in P}{\sum^{\infty}} n Q = \underset{n \notin P}{\sum^{\infty}} n

P = 2 + 3 + 5 + 7 + \dots

Q = 1 + 4 + 6 + 8 + \dots

Is P > Q ? Is Q > P ?

Commented by prakash jain last updated on 11/Jun/17

Let P_{k} be sum of first k prime

numbers

f o r l a r g e k

P_{k} \sim \frac{1}{2} k^{2} \ln k

N_{k} = \frac{k (k + 1)}{2} (sum of first k natural numbers)

Clearly P_{k} > N_{k}

Also P_{k} > Q_{k} .

So \lim_{k \to \infty} (P_{k} - Q_{k}) = \infty

How we cannot say anything about

infinite sum

P - Q = \lim_{k \to \infty} P_{k} - \lim_{k \to \infty} Q_{k} = ?

Commented by FilupS last updated on 11/Jun/17

Commented by FilupS last updated on 11/Jun/17

This is what I found, from playing

around . But it is based off assumption

Commented by linkelly0615 last updated on 12/Jun/17

h m m \dots \dots . w h y

P_{k} ∽ \frac{1}{2} k^{2} l n (k)

Commented by FilupS last updated on 12/Jun/17

Prime number theory

Answered by linkelly0615 last updated on 11/Jun/17

u h \dots \dots

P a n d Q a r e n o t N U M B E R S

s o w e c a n n o t j u d g e w h i c h i s l a r g e r ∽

Commented by FilupS last updated on 11/Jun/17

You can, I think . It {isn}^{'} t uncommon

to show that one infinite sequence is

larger than another .

e . g .

A = \underset{n = 1}{\sum^{\infty}} n B = \underset{n = 1}{\sum^{\infty}} (n + 1)

In this case :

B = \underset{n = 1}{\sum^{\infty}} n + \underset{n = 1}{\sum^{\infty}} 1

B = A + \underset{n = 1}{\sum^{\infty}} 1

∴ B > A

Commented by linkelly0615 last updated on 11/Jun/17

B u t \dots . . i n m y o p i n i o n \dots .

i f

a_{n} = \underset{i = 1}{\sum^{n}} i, b_{m} = \underset{j = 1}{\sum^{m}} (j + 1)

\lim_{m \to \infty} b_{m} - \lim_{n \to \infty} a_{n} = B - A (s a m e c a s e)

\lim_{m \to \infty} b_{m} - \lim_{n \to \infty} a_{n} =

\lim_{m \to \infty} [\underset{j = 1}{\sum^{m}} (j + 1)] - \lim_{n \to \infty} [\underset{i = 1}{\sum^{n}} i] =

\lim_{m \to \infty} [\underset{j = 1}{\sum^{m}} (j)] + \lim_{m \to \infty} (m) - \lim_{n \to \infty} [\underset{i = 1}{\sum^{n}} i] =

= a_{m} - a_{n} + m, a n d m, n t e n d t o \infty

w e c a n n o t b e s u r e a b o u t w h i c h i s t r u e . a_{m} + m > a_{n}, a_{m} + m = a_{n} o r a_{m} + m < a_{n}

Commented by linkelly0615 last updated on 11/Jun/17

B u t,

i f c_{n} = b_{n} - a_{n} \Rightarrow \lim_{n \to \infty} c_{n} > 0

Commented by FilupS last updated on 11/Jun/17

both limit to infinity

\lim_{m \to \infty} \underset{n = 1}{\sum^{m}} n = \lim_{m \to \infty} \underset{n = 1}{\sum^{m}} (n + 1)

but one is a greater in value as it

approaches infinity .

Commented by linkelly0615 last updated on 11/Jun/17

H m m m

I w o n d e r k n o w \dots . .

I n y o u r o p i n i o n \dots . . h o w w o u l d y o u s h o w^{″} G r e a t e r i n {v a l u e}^{″} i n m a t h m a t i c a l w a y \dots .

Commented by FilupS last updated on 11/Jun/17

Depends .

When talking about sets

∣ A ∣>∣ B ∣

with sequences, one value increases

in value at a faster rate, so, naturally

one sequence should be bigger if you were

to stop

Commented by linkelly0615 last updated on 11/Jun/17

t h e t h i n g y o u s a i d i s

^{″} ∣ a_{n} ∣>∣ b_{n} ∣^{″}

Commented by FilupS last updated on 11/Jun/17

∣ A ∣ in set theory, tells you the size of

the set .

e . g .

S = {a_{1}, a_{2}, \dots, a_{n}}

∣ S ∣= n

Commented by FilupS last updated on 11/Jun/17

For sequences,

let A = \underset{n = 1}{\sum^{m}} n, B = \underset{n = 1}{\sum^{m}} (n + 1)

\underline{m ∣ A ∣ B}

1 ∣ 1 ∣ 2

2 ∣ 3 ∣ 5

3 ∣ 7 ∣ 10

⋮

\forall m, B > A

Commented by prakash jain last updated on 11/Jun/17

To Filup

it is not valid to write

\underset{n = 1}{\sum^{\infty}} (n + 1) = \underset{n = 1}{\sum^{\infty}} n + \underset{n = 1}{\sum^{\infty}} 1

The above can only be used

for series which are absolutely

convergent since it involves

reordering .

Also look up Riemann rearrangement

theorem .

Commented by FilupS last updated on 11/Jun/17

Thanks, i will

Commented by FilupS last updated on 11/Jun/17

I see what you mean .

So beacause \underset{n = 1}{\sum^{\infty}} a_{n} doesnt converge,

it cant be rearranged ?

Does my point still hold valid ?

Is B > A ?

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