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=Σ_(n∈P) ^∞ n Q=Σ_(n∉P) ^∞ n P=2+3+5+7+... Q=1+4+6+8+... Is P>Q? Is Q>P?

$${P}=\underset{{n}\in\mathbb{P}} {\overset{\infty} {\sum}}{n}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Q}=\underset{{n}\notin\mathbb{P}} {\overset{\infty} {\sum}}{n} \\ $$$${P}=\mathrm{2}+\mathrm{3}+\mathrm{5}+\mathrm{7}+… \\ $$$${Q}=\mathrm{1}+\mathrm{4}+\mathrm{6}+\mathrm{8}+… \\ $$$$\: \\ $$$$\mathrm{Is}\:{P}>{Q}?\:\:\:\mathrm{Is}\:{Q}>{P}? \\ $$

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

Let P_k be sum of first k prime numbers for large k P_k ∼(1/2)k^2 ln k N_k =((k(k+1))/2) (sum of first k natural numbers) Clearly P_k >N_k Also P_k >Q_k . So lim_(k→∞) (P_k −Q_k )=∞ How we cannot say anything about infinite sum P−Q=lim_(k→∞) P_k −lim_(k→∞) Q_k =?

$$\mathrm{Let}\:\mathrm{P}_{{k}} \:\mathrm{be}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:{k}\:\mathrm{prime} \\ $$$$\mathrm{numbers} \\ $$$${for}\:{large}\:{k} \\ $$$$\mathrm{P}_{{k}} \sim\frac{\mathrm{1}}{\mathrm{2}}{k}^{\mathrm{2}} \mathrm{ln}\:{k} \\ $$$${N}_{{k}} =\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}}\:\left(\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:{k}\:\mathrm{natural}\:\mathrm{numbers}\right) \\ $$$$\mathrm{Clearly}\:{P}_{{k}} >{N}_{{k}} \\ $$$$\mathrm{Also}\:{P}_{{k}} >{Q}_{{k}} . \\ $$$$\mathrm{So}\:\underset{{k}\rightarrow\infty} {\mathrm{lim}}\left({P}_{{k}} −{Q}_{{k}} \right)=\infty \\ $$$$\mathrm{How}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{say}\:\mathrm{anything}\:\mathrm{about} \\ $$$$\mathrm{infinite}\:\mathrm{sum} \\ $$$$\:{P}−{Q}=\underset{{k}\rightarrow\infty} {\mathrm{lim}}{P}_{{k}} −\underset{{k}\rightarrow\infty} {\mathrm{lim}}{Q}_{{k}} =? \\ $$

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

$$\mathrm{This}\:\mathrm{is}\:\mathrm{what}\:\mathrm{I}\:\mathrm{found},\:\mathrm{from}\:\mathrm{playing} \\ $$$$\mathrm{around}.\:\mathrm{But}\:\mathrm{it}\:\mathrm{is}\:\mathrm{based}\:\mathrm{off}\:\mathrm{assumption} \\ $$

Commented by linkelly0615 last updated on 12/Jun/17

$${hmm}…….{why}\: \\ $$$${P}_{{k}} \backsim\frac{\mathrm{1}}{\mathrm{2}}{k}^{\mathrm{2}} {ln}\left({k}\right) \\ $$

Commented by FilupS last updated on 12/Jun/17

$$\mathrm{Prime}\:\mathrm{number}\:\mathrm{theory} \\ $$

Answered by linkelly0615 last updated on 11/Jun/17

$${uh}…… \\ $$$${P}\:{and}\:{Q}\:{are}\:{not}\:{NUMBERS} \\ $$$${so}\:{we}\:{can}\:{not}\:{judge}\:{which}\:{is}\:{larger}\backsim \\ $$

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=Σ_(n=1) ^∞ n B=Σ_(n=1) ^∞ (n+1) In this case: B=Σ_(n=1) ^∞ n+Σ_(n=1) ^∞ 1 B=A+Σ_(n=1) ^∞ 1 ∴B>A

$$\mathrm{You}\:\mathrm{can},\:\mathrm{I}\:\mathrm{think}.\:\mathrm{It}\:\mathrm{isn}'\mathrm{t}\:\mathrm{uncommon} \\ $$$$\mathrm{to}\:\mathrm{show}\:\mathrm{that}\:\mathrm{one}\:\mathrm{infinite}\:\mathrm{sequence}\:\mathrm{is} \\ $$$$\mathrm{larger}\:\mathrm{than}\:\mathrm{another}. \\ $$$$\: \\ $$$$\mathrm{e}.\mathrm{g}. \\ $$$${A}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}\:\:\:\:\:\:\:\:\:\:\:\:{B}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({n}+\mathrm{1}\right) \\ $$$$\mathrm{In}\:\mathrm{this}\:\mathrm{case}: \\ $$$${B}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{1} \\ $$$${B}={A}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{1} \\ $$$$\therefore{B}>{A} \\ $$

Commented by linkelly0615 last updated on 11/Jun/17

But.....in my opinion.... if a_n =Σ_(i=1) ^n i, b_m =Σ_(j=1) ^m (j+1) lim_(m→∞) b_m −lim_(n→∞) a_n =B−A (same case) lim_(m→∞) b_m −lim_(n→∞) a_n = lim_(m→∞) [Σ_(j=1) ^m (j+1)]−lim_(n→∞) [Σ_(i=1) ^n i]= lim_(m→∞) [Σ_(j=1) ^m (j)]+lim_(m→∞) (m)−lim_(n→∞) [Σ_(i=1) ^n i]= =a_m −a_n +m, and m,n tend to ∞ we can not be sure about which is true. a_m +m>a_n , a_m +m=a_n or a_m +m<a_n

$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${But}…..{in}\:{my}\:{opinion}…. \\ $$$${if} \\ $$$${a}_{{n}} =\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i},\:{b}_{{m}} =\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\left({j}+\mathrm{1}\right) \\ $$$$\underset{{m}\rightarrow\infty} {\mathrm{lim}}{b}_{{m}} −\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} ={B}−{A}\:\left({same}\:{case}\right) \\ $$$$\underset{{m}\rightarrow\infty} {\mathrm{lim}}{b}_{{m}} −\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} = \\ $$$$\underset{{m}\rightarrow\infty} {\mathrm{lim}}\left[\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\left({j}+\mathrm{1}\right)\right]−\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}\right]= \\ $$$$\underset{{m}\rightarrow\infty} {\mathrm{lim}}\left[\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\left({j}\right)\right]+\underset{{m}\rightarrow\infty} {\mathrm{lim}}\left({m}\right)−\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}\right]= \\ $$$$={a}_{{m}} −{a}_{{n}} +{m},\:{and}\:{m},{n}\:{tend}\:{to}\:\infty \\ $$$${we}\:{can}\:{not}\:{be}\:{sure}\:{about}\:{which}\:{is}\:{true}.\:{a}_{{m}} +{m}>{a}_{{n}} \:,\:{a}_{{m}} +{m}={a}_{{n}} \:{or}\:\:{a}_{{m}} +{m}<{a}_{{n}} \: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by linkelly0615 last updated on 11/Jun/17

$${But}, \\ $$$${if}\:\:\:\:\:\:{c}_{{n}} ={b}_{{n}} −{a}_{{n}} \:\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}{c}_{{n}} >\mathrm{0} \\ $$

Commented by FilupS last updated on 11/Jun/17

both limit to infinity lim_(m→∞) Σ_(n=1) ^m n=lim_(m→∞) Σ_(n=1) ^m (n+1) but one is a greater in value as it approaches infinity.

$$\mathrm{both}\:\mathrm{limit}\:\mathrm{to}\:\mathrm{infinity} \\ $$$$\underset{{m}\rightarrow\infty} {\mathrm{lim}}\:\underset{{n}=\mathrm{1}} {\overset{{m}} {\sum}}{n}=\underset{{m}\rightarrow\infty} {\mathrm{lim}}\underset{{n}=\mathrm{1}} {\overset{{m}} {\sum}}\left({n}+\mathrm{1}\right) \\ $$$$\mathrm{but}\:\mathrm{one}\:\mathrm{is}\:\mathrm{a}\:\mathrm{greater}\:\mathrm{in}\:\mathrm{value}\:\mathrm{as}\:\mathrm{it} \\ $$$$\mathrm{approaches}\:\mathrm{infinity}. \\ $$

Commented by linkelly0615 last updated on 11/Jun/17

$${Hmmm} \\ $$$${I}\:{wonder}\:{know}\:….. \\ $$$${In}\:{your}\:{opinion}…..{how}\:{would}\:{you}\:{show}\:''{Greater}\:{in}\:{value}''\:{in}\:{mathmatical}\:{way}…. \\ $$

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

$$\mathrm{Depends}. \\ $$$$\mathrm{When}\:\mathrm{talking}\:\mathrm{about}\:\mathrm{sets} \\ $$$$\mid{A}\mid>\mid{B}\mid \\ $$$$\: \\ $$$$\mathrm{with}\:\mathrm{sequences},\:\mathrm{one}\:\mathrm{value}\:\mathrm{increases} \\ $$$$\mathrm{in}\:\mathrm{value}\:\mathrm{at}\:\mathrm{a}\:\mathrm{faster}\:\mathrm{rate},\:\mathrm{so},\:\mathrm{naturally} \\ $$$$\mathrm{one}\:\mathrm{sequence}\:\mathrm{should}\:\mathrm{be}\:\mathrm{bigger}\:\mathrm{if}\:\mathrm{you}\:\mathrm{were} \\ $$$$\mathrm{to}\:\mathrm{stop} \\ $$

Commented by linkelly0615 last updated on 11/Jun/17

$${the}\:{thing}\:{you}\:{said}\:{is} \\ $$$$''\mid{a}_{{n}} \mid>\mid{b}_{{n}} \mid'' \\ $$

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 , ..., a_n } ∣S∣=n

$$\mid{A}\mid\:\mathrm{in}\:\mathrm{set}\:\mathrm{theory},\:\mathrm{tells}\:\mathrm{you}\:\mathrm{the}\:\mathrm{size}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{set}. \\ $$$$\mathrm{e}.\mathrm{g}. \\ $$$${S}=\left\{{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:…,\:{a}_{{n}} \right\} \\ $$$$\mid{S}\mid={n} \\ $$

Commented by FilupS last updated on 11/Jun/17

For sequences, let A=Σ_(n=1) ^m n, B=Σ_(n=1) ^m (n+1) m ∣ A ∣ B_(−) 1 ∣ 1 ∣ 2 2 ∣ 3 ∣ 5 3 ∣ 7 ∣ 10 ⋮ ∀m, B>A

$$\mathrm{For}\:\mathrm{sequences}, \\ $$$$\mathrm{let}\:{A}=\underset{{n}=\mathrm{1}} {\overset{{m}} {\sum}}{n},\:\:{B}=\underset{{n}=\mathrm{1}} {\overset{{m}} {\sum}}\left({n}+\mathrm{1}\right) \\ $$$$\underset{−} {{m}\:\mid\:{A}\:\mid\:{B}} \\ $$$$\mathrm{1}\:\:\:\mid\:\:\mathrm{1}\:\:\mid\:\:\mathrm{2} \\ $$$$\mathrm{2}\:\:\:\mid\:\:\mathrm{3}\:\:\mid\:\:\mathrm{5} \\ $$$$\mathrm{3}\:\:\:\mid\:\:\mathrm{7}\:\:\mid\:\:\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\vdots \\ $$$$\forall{m},\:{B}>{A} \\ $$

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

To Filup it is not valid to write Σ_(n=1) ^∞ (n+1)=Σ_(n=1) ^∞ n+Σ_(n=1) ^∞ 1 The above can only be used for series which are absolutely convergent since it involves reordering. Also look up Riemann rearrangement theorem.

$$\mathrm{To}\:\mathrm{Filup} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{valid}\:\mathrm{to}\:\mathrm{write} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({n}+\mathrm{1}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{1} \\ $$$$\mathrm{The}\:\mathrm{above}\:\mathrm{can}\:\mathrm{only}\:\mathrm{be}\:\mathrm{used} \\ $$$$\mathrm{for}\:\mathrm{series}\:\mathrm{which}\:\mathrm{are}\:\mathrm{absolutely} \\ $$$$\mathrm{convergent}\:\mathrm{since}\:\mathrm{it}\:\mathrm{involves} \\ $$$$\mathrm{reordering}. \\ $$$$\mathrm{Also}\:\mathrm{look}\:\mathrm{up}\:\mathrm{Riemann}\:\mathrm{rearrangement} \\ $$$$\mathrm{theorem}. \\ $$

Commented by FilupS last updated on 11/Jun/17

$$\mathrm{Thanks},\:\mathrm{i}\:\mathrm{will} \\ $$

Commented by FilupS last updated on 11/Jun/17

I see what you mean. So beacause Σ_(n=1) ^∞ a_n doesnt converge, it cant be rearranged? Does my point still hold valid? Is B>A?

$$\mathrm{I}\:\mathrm{see}\:\mathrm{what}\:\mathrm{you}\:\mathrm{mean}. \\ $$$$\mathrm{So}\:\mathrm{beacause}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} \:\mathrm{doesnt}\:\mathrm{converge}, \\ $$$$\mathrm{it}\:\mathrm{cant}\:\mathrm{be}\:\mathrm{rearranged}? \\ $$$$\: \\ $$$$\mathrm{Does}\:\mathrm{my}\:\mathrm{point}\:\mathrm{still}\:\mathrm{hold}\:\mathrm{valid}? \\ $$$$\mathrm{Is}\:{B}>{A}? \\ $$

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