Question Number 219098 by zetamaths last updated on 20/Apr/25
$$\zeta\left(\alpha\right)=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\alpha} }\:\: \\ $$
Answered by Frix last updated on 20/Apr/25
$$\mathrm{This}\:\mathrm{is}\:\mathrm{just}\:\mathrm{the}\:\mathrm{definition}\:\mathrm{of}\:\mathrm{the}\:\zeta−\mathrm{Function}, \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{nothing}\:\mathrm{to}\:\mathrm{solve}. \\ $$$$\forall{s}\in\mathbb{C}\wedge\mathrm{Re}\left({s}\right)>\mathrm{1}:\:\zeta\left({s}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{{s}} } \\ $$
Answered by MrGaster last updated on 20/Apr/25
![∫_1 ^∞ (1/x^α )dx=lim_(t→∞) [(x^(−α+1) /(−α+1))]_1 ^t =lim_(t→∞) ((t^(−α+1) /(−α+1))−(1/(−α+1))) ⇒(α>1)⇒(−α+1≤0)lim_(t→∞) t^(−α+1) =0 ⇒∫_1 ^∞ (1/x^α )dx=(1/(α−1))<∞ ⇒Σ_(n=1) ^∞ (1/n^α )_(converges) (α≤1)⇒∫_1 ^∞ (1/x^α )dx_(diverges) ⇒Σ_(n=1) ^∞ (1/n^α )diverges ∴α>1](https://www.tinkutara.com/question/Q219184.png)
$$\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{x}^{\alpha} }{dx}=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\left[\frac{{x}^{−\alpha+\mathrm{1}} }{−\alpha+\mathrm{1}}\right]_{\mathrm{1}} ^{{t}} \\ $$$$=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{t}^{−\alpha+\mathrm{1}} }{−\alpha+\mathrm{1}}−\frac{\mathrm{1}}{−\alpha+\mathrm{1}}\right) \\ $$$$\Rightarrow\left(\alpha>\mathrm{1}\right)\Rightarrow\left(−\alpha+\mathrm{1}\leqslant\mathrm{0}\right)\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:{t}^{−\alpha+\mathrm{1}} =\mathrm{0} \\ $$$$\Rightarrow\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{x}^{\alpha} }{dx}=\frac{\mathrm{1}}{\alpha−\mathrm{1}}<\infty \\ $$$$\Rightarrow\underset{\mathrm{converges}} {\underbrace{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\alpha} }}} \\ $$$$\left(\alpha\leq\mathrm{1}\right)\Rightarrow\underset{\mathrm{diverges}} {\underbrace{\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{x}^{\alpha} }{dx}}}\:\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\alpha} }\mathrm{diverges} \\ $$$$\therefore\alpha>\mathrm{1} \\ $$
Commented by zetamaths last updated on 20/Apr/25
$${is}\:{false}\:{use}\:{gamma}\:{function} \\ $$