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AllQuestion and Answers: Page 62
Question Number 219098 Answers: 2 Comments: 0
$$\zeta\left(\alpha\right)=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\alpha} }\:\: \\ $$
Question Number 219093 Answers: 3 Comments: 1
Question Number 219090 Answers: 2 Comments: 0
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequence}\:{a}_{{n}} =\frac{\mathrm{1}}{\:\sqrt[{{n}}]{{n}!}}\:\mathrm{is}\:\mathrm{decreasing}. \\ $$
Question Number 219088 Answers: 0 Comments: 0
Question Number 219087 Answers: 1 Comments: 0
Question Number 219086 Answers: 2 Comments: 0
Question Number 219085 Answers: 0 Comments: 0
Question Number 219084 Answers: 0 Comments: 0
Question Number 219083 Answers: 0 Comments: 0
Question Number 219078 Answers: 1 Comments: 0
$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{m}}\right).\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \centerdot{k}\centerdot\frac{\left({m}!\right)^{\mathrm{2}} }{\left({m}−{k}\right)!\left({m}+{k}\right)!}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:{Proof}\:{this}\:{formula} \\ $$
Question Number 219077 Answers: 0 Comments: 0
$$\int_{\mathrm{0}} ^{+\infty} \left(\frac{{sin}\left({n}\right)}{{n}}\right)^{{m}} {dn}=\pi\centerdot\frac{{m}}{\mathrm{2}^{{m}} }\centerdot\underset{\phi=\mathrm{0}} {\overset{{m}/\mathrm{2}} {\sum}}\left(−\mathrm{1}\right)^{\emptyset} \centerdot\frac{\left({n}−\mathrm{2}\phi\right)^{{m}−\mathrm{1}} }{\left({m}−\phi\right)!\centerdot\phi!}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Proof}\:{this}\:{formula} \\ $$
Question Number 219076 Answers: 0 Comments: 0
Question Number 219071 Answers: 3 Comments: 0
Question Number 219070 Answers: 0 Comments: 0
Question Number 219069 Answers: 0 Comments: 0
Question Number 219068 Answers: 2 Comments: 0
Question Number 219067 Answers: 2 Comments: 1
Question Number 219066 Answers: 4 Comments: 0
Question Number 219065 Answers: 3 Comments: 0
Question Number 219060 Answers: 2 Comments: 0
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{{m}} }{{x}^{{n}} }{dx},{n}\in\mathbb{N},{m}\in\mathbb{N},{n}\leqslant{m} \\ $$
Question Number 219135 Answers: 1 Comments: 0
Question Number 219025 Answers: 3 Comments: 0
Question Number 219004 Answers: 0 Comments: 0
Question Number 219003 Answers: 1 Comments: 0
Question Number 222659 Answers: 1 Comments: 0
Question Number 218970 Answers: 4 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2},\mathrm{12},\mathrm{18},\mathrm{48},\mathrm{50},..... \\ $$$$ \\ $$
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