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Question Number 206858 by Ghisom last updated on 27/Apr/24

$$\mathrm{prove}\mathrm{that}$$$$H_n=\underset{0}{\stackrel{1}{\int }}\frac{t^n-1}{t-1}dt$$

Answered by mathzup last updated on 28/Apr/24

$$I_{\xi }={\int }_{\xi }^1\frac{t^n-1}{t-1}dtwehave{H}_n={lim}_{\xi \to 0^{+}}{I}_{\xi }$$$$but{I}_{\xi }={\int }_{\xi }^1\frac{(t-1)(1+t+t^2+....+t^{n-1})}{t-1}dt$$$$={\int }_{\xi }^1(1+t+t^2+....+t^{n-1})dt$$$$={[t+\frac{t^2}{2}+\frac{t^3}{3}+....+\frac{t^n}{n}]}_{\xi }^1$$$$=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}-(\xi +\frac{{\xi }^2}{2}+....+\frac{{\xi }^n}{n})$$$$\Rightarrow {lim}_{\xi \to 0}{I}_{\xi }=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}=H_n$$

Commented by Ghisom last updated on 28/Apr/24

thank you

Answered by JDamian last updated on 28/Apr/24

$$\frac{t^n-1}{t-1}=1+t+t^2+\cdot \cdot \cdot +t^{n-1}G.P.$$$$\underset{0}{\stackrel{1}{\int }}1+t+t^2+\cdot \cdot \cdot +t^{n-1}dt=$$$$={[t+\frac{t^2}{2}+\frac{t^3}{3}+\cdot \cdot \cdot +\frac{t^n}{n}]}_0^1=$$$$=(1+\frac{1}{2}+\frac{1}{3}+\cdot \cdot \cdot +\frac{1}{n})-(0+0+\cdot \cdot \cdot +0)=$$$$=1+\frac{1}{2}+\frac{1}{3}+\cdot \cdot \cdot +\frac{1}{n}={\mathrm{H}}_n◼$$

Commented by Ghisom last updated on 28/Apr/24

thank you

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