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Question Number 155399 by 0731619 last updated on 30/Sep/21

Answered by TheHoneyCat last updated on 30/Sep/21

$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\underset{k=0}{\sum ^n}\frac{1}{k}=+\mathrm{\infty }$$$$\mathrm{hence}\frac{1}{\mathrm{\Sigma }\frac{1}{n}}\underset{n\to \mathrm{\infty }}{\to }0$$

Answered by Niiicooooo last updated on 30/Sep/21

$$\underset{k=1}{\sum ^n}\frac{1}{k}\sim lnn+\gamma (n\to \mathrm{\infty })$$

Commented by puissant last updated on 30/Sep/21

$$\underset{k=1}{\sum ^n}\frac{1}{k}=lnn+\gamma +\underset{n\mathrm{\infty }}{o}\left(1\right)$$

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