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Question Number 3062 by 123456 last updated on 04/Dec/15

ω(z) (Q3035) does ω(z)=0 for z∈N^∗ ?

$$\omega\left({z}\right)\:\left(\mathrm{Q3035}\right) \\ $$$$\mathrm{does} \\ $$$$\omega\left({z}\right)=\mathrm{0}\:\mathrm{for}\:{z}\in\mathbb{N}^{\ast} ? \\ $$

Commented by Filup last updated on 04/Dec/15

What is N^∗ ?

$$\mathrm{What}\:\mathrm{is}\:\mathbb{N}^{\ast} ? \\ $$

Commented by Filup last updated on 04/Dec/15

ω(z)=lim_(n→∞) ((n^z [z+(z+1)+...+(z+n)])/(z(z+1)∙∙∙(z+n)))

$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}} \left[{z}+\left({z}+\mathrm{1}\right)+...+\left({z}+{n}\right)\right]}{{z}\left({z}+\mathrm{1}\right)\centerdot\centerdot\centerdot\left({z}+{n}\right)} \\ $$

Commented by 123456 last updated on 04/Dec/15

N^∗ =N/{0}

$$\mathbb{N}^{\ast} =\mathbb{N}/\left\{\mathrm{0}\right\} \\ $$

Commented by Filup last updated on 04/Dec/15

N/{0}={±1, ±2, ±3} correct? sorry, i have forgotten this set :P

$$\mathbb{N}/\left\{\mathrm{0}\right\}=\left\{\pm\mathrm{1},\:\pm\mathrm{2},\:\pm\mathrm{3}\right\} \\ $$$$\mathrm{correct}?\:\mathrm{sorry},\:\mathrm{i}\:\mathrm{have}\:\mathrm{forgotten}\:\mathrm{this}\:\mathrm{set} \\ $$$$:\mathrm{P} \\ $$

Commented by 123456 last updated on 04/Dec/15

N={0,1,2,...} (natural numbers) the set you shown is Z={0,±1,±2,...} (integer numbers) Z^∗ =Z/{0}={±1,±2,...}

$$\mathbb{N}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},...\right\}\:\left(\mathrm{natural}\:\mathrm{numbers}\right) \\ $$$$\mathrm{the}\:\mathrm{set}\:\mathrm{you}\:\mathrm{shown}\:\mathrm{is} \\ $$$$\mathbb{Z}=\left\{\mathrm{0},\pm\mathrm{1},\pm\mathrm{2},...\right\}\:\left(\mathrm{integer}\:\mathrm{numbers}\right) \\ $$$$\mathbb{Z}^{\ast} =\mathbb{Z}/\left\{\mathrm{0}\right\}=\left\{\pm\mathrm{1},\pm\mathrm{2},...\right\} \\ $$

Commented by Filup last updated on 04/Dec/15

Ah! Thanks!!!

$${Ah}!\:\mathrm{T}{hanks}!!! \\ $$

Commented by prakash jain last updated on 04/Dec/15

There isn′t any accepted convention for natural numbers so Z^+ ={1,2,3,4,...} is used.

$$\mathrm{There}\:\mathrm{isn}'\mathrm{t}\:\mathrm{any}\:\mathrm{accepted}\:\mathrm{convention}\:\mathrm{for}\: \\ $$$$\mathrm{natural}\:\mathrm{numbers}\:\mathrm{so}\:\mathbb{Z}^{+} \:=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},...\right\}\:\mathrm{is}\:\mathrm{used}. \\ $$

Commented by 123456 last updated on 04/Dec/15

its have N={0,1,2,...} or N={1,2,3,...} its depends from people to people and source to source, the two have their advantagens

$$\mathrm{its}\:\mathrm{have} \\ $$$$\mathbb{N}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},...\right\} \\ $$$$\mathrm{or} \\ $$$$\mathbb{N}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},...\right\} \\ $$$$\mathrm{its}\:\mathrm{depends}\:\mathrm{from}\:\mathrm{people}\:\mathrm{to}\:\mathrm{people}\:\mathrm{and} \\ $$$$\mathrm{source}\:\mathrm{to}\:\mathrm{source},\:\mathrm{the}\:\mathrm{two}\:\mathrm{have}\:\mathrm{their} \\ $$$$\mathrm{advantagens} \\ $$$$ \\ $$

Answered by Filup last updated on 04/Dec/15

ω(z)=lim_(n→∞) ((n^z [z+(z+1)+...+(z+n)])/(z(z+1)∙∙∙(z+n))) ω(z)=lim_(n→∞) ((n^z (z+Σ_(i=1) ^n (z+i)))/(((z+n)!)/((z−1)!))) ω(z)=lim_(n→∞) ((n^z [z+(1/2)n(2z+n+1)])/(((z+n)!)/((z−1)!))) ω(z)=lim_(n→∞) ((n^z [z+(1/2)(2zn+n^2 +n)])/(((z+n)!)/((z−1)!))) ω(z)=lim_(n→∞) (([2zn^z +2zn^(z+1) +n^(z+2) +n^(z+1) ](z−1)!)/(2(z+n)!)) ω(z)=lim_(n→∞) (((2zn^z +2z(n^(z+1) +1)+n^(z+2) )(z−1)!)/(2(z+n)!)) ω(z)=lim_(n→∞) (((2z(n^z +n^(z+1) +1)+n^(z+2) )(z−1)!)/(2(z+n)!)) ω(z)=lim_(n→∞) ((2z(n^z +n^(z+1) +1)+n^(z+2) )/(2(z+n)!)) (z−1)! ω(z)=lim_(n→∞) (((z(n^z +n^(z+1) +1))/((z+n)!)) + (n^(z+2) /(2(z+n)!)))(z−1)! ω(z)=lim_(n→∞) (z−1)!∙(z[(n^z /((z+n)!))+(n^(z+1) /((z+n)!))+(1/((z+n)!))] + (n^(z+2) /(2(z+n)!))) continue (will edit if I figure it out)

$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}} \left[{z}+\left({z}+\mathrm{1}\right)+...+\left({z}+{n}\right)\right]}{{z}\left({z}+\mathrm{1}\right)\centerdot\centerdot\centerdot\left({z}+{n}\right)} \\ $$$$ \\ $$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}} \left({z}+\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({z}+{i}\right)\right)}{\frac{\left({z}+{n}\right)!}{\left({z}−\mathrm{1}\right)!}} \\ $$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}} \left[{z}+\frac{\mathrm{1}}{\mathrm{2}}{n}\left(\mathrm{2}{z}+{n}+\mathrm{1}\right)\right]}{\frac{\left({z}+{n}\right)!}{\left({z}−\mathrm{1}\right)!}} \\ $$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}} \left[{z}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{zn}+{n}^{\mathrm{2}} +{n}\right)\right]}{\frac{\left({z}+{n}\right)!}{\left({z}−\mathrm{1}\right)!}} \\ $$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left[\mathrm{2}{zn}^{{z}} +\mathrm{2}{zn}^{{z}+\mathrm{1}} +{n}^{{z}+\mathrm{2}} +{n}^{{z}+\mathrm{1}} \right]\left({z}−\mathrm{1}\right)!}{\mathrm{2}\left({z}+{n}\right)!} \\ $$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\mathrm{2}{zn}^{{z}} +\mathrm{2}{z}\left({n}^{{z}+\mathrm{1}} +\mathrm{1}\right)+{n}^{{z}+\mathrm{2}} \right)\left({z}−\mathrm{1}\right)!}{\mathrm{2}\left({z}+{n}\right)!} \\ $$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\mathrm{2}{z}\left({n}^{{z}} +{n}^{{z}+\mathrm{1}} +\mathrm{1}\right)+{n}^{{z}+\mathrm{2}} \right)\left({z}−\mathrm{1}\right)!}{\mathrm{2}\left({z}+{n}\right)!} \\ $$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}{z}\left({n}^{{z}} +{n}^{{z}+\mathrm{1}} +\mathrm{1}\right)+{n}^{{z}+\mathrm{2}} }{\mathrm{2}\left({z}+{n}\right)!}\:\left({z}−\mathrm{1}\right)! \\ $$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{z}\left({n}^{{z}} +{n}^{{z}+\mathrm{1}} +\mathrm{1}\right)}{\left({z}+{n}\right)!}\:+\:\frac{{n}^{{z}+\mathrm{2}} }{\mathrm{2}\left({z}+{n}\right)!}\right)\left({z}−\mathrm{1}\right)! \\ $$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left({z}−\mathrm{1}\right)!\centerdot\left({z}\left[\frac{{n}^{{z}} }{\left({z}+{n}\right)!}+\frac{{n}^{{z}+\mathrm{1}} }{\left({z}+{n}\right)!}+\frac{\mathrm{1}}{\left({z}+{n}\right)!}\right]\:+\:\frac{{n}^{{z}+\mathrm{2}} }{\mathrm{2}\left({z}+{n}\right)!}\right) \\ $$$$ \\ $$$$\mathrm{continue}\:\:\left(\mathrm{will}\:\mathrm{edit}\:\mathrm{if}\:\mathrm{I}\:\mathrm{figure}\:\mathrm{it}\:\mathrm{out}\right) \\ $$

Answered by prakash jain last updated on 04/Dec/15

w_n (z)=((n^z [z+(z+1)+..+(z+n)])/(z(z+1)..(z+n))) =((n^z [((n+1)/2)(2z+n)])/(((z+n)!)/((z−1)!))) =((n^z (z−1)!(n+1)(2z+n))/(2(z+n)!)) =(((z−1)!)/2)∙((n^z (2nz+2z+n^2 ))/((n+z)!)) =k_1 ∙(n^(z+1) /((n+z)!))+k_2 ∙(n^z /((n+z)!))+k_3 ∙(n^(z+2) /((n+z)!)) lim_(n→∞) (n^(z+2) /((n+z)!))=lim_(n→∞) (n^(z+2) /((n+z)!))≤lim_(n→∞) (n^(z+2) /((n−1)!n^(z+1) )) =lim_(n→∞) (n/((n−1)!))=0 simlarly lim_(n→∞) (n^(z+1) /((n+z)!))=0 and lim_(n→∞) (n^z /((n+z)!))=0 lim_(n→∞) w_n (z)=w(z)=0

$${w}_{{n}} \left({z}\right)=\frac{{n}^{{z}} \left[{z}+\left({z}+\mathrm{1}\right)+..+\left({z}+{n}\right)\right]}{{z}\left({z}+\mathrm{1}\right)..\left({z}+{n}\right)} \\ $$$$=\frac{{n}^{{z}} \left[\frac{{n}+\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{z}+{n}\right)\right]}{\frac{\left({z}+{n}\right)!}{\left({z}−\mathrm{1}\right)!}} \\ $$$$=\frac{{n}^{{z}} \left({z}−\mathrm{1}\right)!\left({n}+\mathrm{1}\right)\left(\mathrm{2}{z}+{n}\right)}{\mathrm{2}\left({z}+{n}\right)!} \\ $$$$=\frac{\left({z}−\mathrm{1}\right)!}{\mathrm{2}}\centerdot\frac{{n}^{{z}} \left(\mathrm{2}{nz}+\mathrm{2}{z}+{n}^{\mathrm{2}} \right)}{\left({n}+{z}\right)!} \\ $$$$={k}_{\mathrm{1}} \centerdot\frac{{n}^{{z}+\mathrm{1}} }{\left({n}+{z}\right)!}+{k}_{\mathrm{2}} \centerdot\frac{{n}^{{z}} }{\left({n}+{z}\right)!}+{k}_{\mathrm{3}} \centerdot\frac{{n}^{{z}+\mathrm{2}} }{\left({n}+{z}\right)!} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}^{{z}+\mathrm{2}} }{\left({n}+{z}\right)!}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}+\mathrm{2}} }{\left({n}+{z}\right)!}\leqslant\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}+\mathrm{2}} }{\left({n}−\mathrm{1}\right)!{n}^{{z}+\mathrm{1}} } \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}}{\left({n}−\mathrm{1}\right)!}=\mathrm{0} \\ $$$$\mathrm{simlarly}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}^{{z}+\mathrm{1}} }{\left({n}+{z}\right)!}=\mathrm{0} \\ $$$$\mathrm{and}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}^{{z}} }{\left({n}+{z}\right)!}=\mathrm{0} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{w}_{{n}} \left({z}\right)={w}\left({z}\right)=\mathrm{0} \\ $$

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