# z-Q3035-does-z-0-for-z-N-

Question Number 3062 by 123456 last updated on 04/Dec/15
$$\omega\left({z}\right)\:\left(\mathrm{Q3035}\right) \\$$$$\mathrm{does} \\$$$$\omega\left({z}\right)=\mathrm{0}\:\mathrm{for}\:{z}\in\mathbb{N}^{\ast} ? \\$$
Commented by Filup last updated on 04/Dec/15
$$\mathrm{What}\:\mathrm{is}\:\mathbb{N}^{\ast} ? \\$$
Commented by Filup last updated on 04/Dec/15
$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}} \left[{z}+\left({z}+\mathrm{1}\right)+…+\left({z}+{n}\right)\right]}{{z}\left({z}+\mathrm{1}\right)\centerdot\centerdot\centerdot\left({z}+{n}\right)} \\$$
Commented by 123456 last updated on 04/Dec/15
$$\mathbb{N}^{\ast} =\mathbb{N}/\left\{\mathrm{0}\right\} \\$$
Commented by Filup last updated on 04/Dec/15
$$\mathbb{N}/\left\{\mathrm{0}\right\}=\left\{\pm\mathrm{1},\:\pm\mathrm{2},\:\pm\mathrm{3}\right\} \\$$$$\mathrm{correct}?\:\mathrm{sorry},\:\mathrm{i}\:\mathrm{have}\:\mathrm{forgotten}\:\mathrm{this}\:\mathrm{set} \\$$$$:\mathrm{P} \\$$
Commented by 123456 last updated on 04/Dec/15
$$\mathbb{N}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},…\right\}\:\left(\mathrm{natural}\:\mathrm{numbers}\right) \\$$$$\mathrm{the}\:\mathrm{set}\:\mathrm{you}\:\mathrm{shown}\:\mathrm{is} \\$$$$\mathbb{Z}=\left\{\mathrm{0},\pm\mathrm{1},\pm\mathrm{2},…\right\}\:\left(\mathrm{integer}\:\mathrm{numbers}\right) \\$$$$\mathbb{Z}^{\ast} =\mathbb{Z}/\left\{\mathrm{0}\right\}=\left\{\pm\mathrm{1},\pm\mathrm{2},…\right\} \\$$
Commented by Filup last updated on 04/Dec/15
$${Ah}!\:\mathrm{T}{hanks}!!! \\$$
Commented by prakash jain last updated on 04/Dec/15
$$\mathrm{There}\:\mathrm{isn}'\mathrm{t}\:\mathrm{any}\:\mathrm{accepted}\:\mathrm{convention}\:\mathrm{for}\: \\$$$$\mathrm{natural}\:\mathrm{numbers}\:\mathrm{so}\:\mathbb{Z}^{+} \:=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},…\right\}\:\mathrm{is}\:\mathrm{used}. \\$$
Commented by 123456 last updated on 04/Dec/15
$$\mathrm{its}\:\mathrm{have} \\$$$$\mathbb{N}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},…\right\} \\$$$$\mathrm{or} \\$$$$\mathbb{N}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},…\right\} \\$$$$\mathrm{its}\:\mathrm{depends}\:\mathrm{from}\:\mathrm{people}\:\mathrm{to}\:\mathrm{people}\:\mathrm{and} \\$$$$\mathrm{source}\:\mathrm{to}\:\mathrm{source},\:\mathrm{the}\:\mathrm{two}\:\mathrm{have}\:\mathrm{their} \\$$$$\mathrm{advantagens} \\$$$$\\$$
Answered by Filup last updated on 04/Dec/15
$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}} \left[{z}+\left({z}+\mathrm{1}\right)+…+\left({z}+{n}\right)\right]}{{z}\left({z}+\mathrm{1}\right)\centerdot\centerdot\centerdot\left({z}+{n}\right)} \\$$$$\\$$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}} \left({z}+\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({z}+{i}\right)\right)}{\frac{\left({z}+{n}\right)!}{\left({z}−\mathrm{1}\right)!}} \\$$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}} \left[{z}+\frac{\mathrm{1}}{\mathrm{2}}{n}\left(\mathrm{2}{z}+{n}+\mathrm{1}\right)\right]}{\frac{\left({z}+{n}\right)!}{\left({z}−\mathrm{1}\right)!}} \\$$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}} \left[{z}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{zn}+{n}^{\mathrm{2}} +{n}\right)\right]}{\frac{\left({z}+{n}\right)!}{\left({z}−\mathrm{1}\right)!}} \\$$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left[\mathrm{2}{zn}^{{z}} +\mathrm{2}{zn}^{{z}+\mathrm{1}} +{n}^{{z}+\mathrm{2}} +{n}^{{z}+\mathrm{1}} \right]\left({z}−\mathrm{1}\right)!}{\mathrm{2}\left({z}+{n}\right)!} \\$$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\mathrm{2}{zn}^{{z}} +\mathrm{2}{z}\left({n}^{{z}+\mathrm{1}} +\mathrm{1}\right)+{n}^{{z}+\mathrm{2}} \right)\left({z}−\mathrm{1}\right)!}{\mathrm{2}\left({z}+{n}\right)!} \\$$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\mathrm{2}{z}\left({n}^{{z}} +{n}^{{z}+\mathrm{1}} +\mathrm{1}\right)+{n}^{{z}+\mathrm{2}} \right)\left({z}−\mathrm{1}\right)!}{\mathrm{2}\left({z}+{n}\right)!} \\$$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}{z}\left({n}^{{z}} +{n}^{{z}+\mathrm{1}} +\mathrm{1}\right)+{n}^{{z}+\mathrm{2}} }{\mathrm{2}\left({z}+{n}\right)!}\:\left({z}−\mathrm{1}\right)! \\$$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{z}\left({n}^{{z}} +{n}^{{z}+\mathrm{1}} +\mathrm{1}\right)}{\left({z}+{n}\right)!}\:+\:\frac{{n}^{{z}+\mathrm{2}} }{\mathrm{2}\left({z}+{n}\right)!}\right)\left({z}−\mathrm{1}\right)! \\$$$$\omega\left({z}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left({z}−\mathrm{1}\right)!\centerdot\left({z}\left[\frac{{n}^{{z}} }{\left({z}+{n}\right)!}+\frac{{n}^{{z}+\mathrm{1}} }{\left({z}+{n}\right)!}+\frac{\mathrm{1}}{\left({z}+{n}\right)!}\right]\:+\:\frac{{n}^{{z}+\mathrm{2}} }{\mathrm{2}\left({z}+{n}\right)!}\right) \\$$$$\\$$$$\mathrm{continue}\:\:\left(\mathrm{will}\:\mathrm{edit}\:\mathrm{if}\:\mathrm{I}\:\mathrm{figure}\:\mathrm{it}\:\mathrm{out}\right) \\$$
Answered by prakash jain last updated on 04/Dec/15
$${w}_{{n}} \left({z}\right)=\frac{{n}^{{z}} \left[{z}+\left({z}+\mathrm{1}\right)+..+\left({z}+{n}\right)\right]}{{z}\left({z}+\mathrm{1}\right)..\left({z}+{n}\right)} \\$$$$=\frac{{n}^{{z}} \left[\frac{{n}+\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{z}+{n}\right)\right]}{\frac{\left({z}+{n}\right)!}{\left({z}−\mathrm{1}\right)!}} \\$$$$=\frac{{n}^{{z}} \left({z}−\mathrm{1}\right)!\left({n}+\mathrm{1}\right)\left(\mathrm{2}{z}+{n}\right)}{\mathrm{2}\left({z}+{n}\right)!} \\$$$$=\frac{\left({z}−\mathrm{1}\right)!}{\mathrm{2}}\centerdot\frac{{n}^{{z}} \left(\mathrm{2}{nz}+\mathrm{2}{z}+{n}^{\mathrm{2}} \right)}{\left({n}+{z}\right)!} \\$$$$={k}_{\mathrm{1}} \centerdot\frac{{n}^{{z}+\mathrm{1}} }{\left({n}+{z}\right)!}+{k}_{\mathrm{2}} \centerdot\frac{{n}^{{z}} }{\left({n}+{z}\right)!}+{k}_{\mathrm{3}} \centerdot\frac{{n}^{{z}+\mathrm{2}} }{\left({n}+{z}\right)!} \\$$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}^{{z}+\mathrm{2}} }{\left({n}+{z}\right)!}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}+\mathrm{2}} }{\left({n}+{z}\right)!}\leqslant\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{z}+\mathrm{2}} }{\left({n}−\mathrm{1}\right)!{n}^{{z}+\mathrm{1}} } \\$$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}}{\left({n}−\mathrm{1}\right)!}=\mathrm{0} \\$$$$\mathrm{simlarly}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}^{{z}+\mathrm{1}} }{\left({n}+{z}\right)!}=\mathrm{0} \\$$$$\mathrm{and}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}^{{z}} }{\left({n}+{z}\right)!}=\mathrm{0} \\$$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{w}_{{n}} \left({z}\right)={w}\left({z}\right)=\mathrm{0} \\$$