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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_(n→+∞ ) Σ_(k=0) ^n (1/k)=+∞  hence (1/(Σ(1/n))) →_(n→∞) 0

$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty\:} \underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}=+\infty \\ $$$$\mathrm{hence}\:\frac{\mathrm{1}}{\Sigma\frac{\mathrm{1}}{{n}}}\:\underset{{n}\rightarrow\infty} {\rightarrow}\mathrm{0} \\ $$

Answered by Niiicooooo last updated on 30/Sep/21

Σ_(k=1) ^n (1/k) ∼ ln n + γ (n→∞)

$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:\sim\:{ln}\:{n}\:+\:\gamma\:\left({n}\rightarrow\infty\right) \\ $$

Commented by puissant last updated on 30/Sep/21

Σ_(k=1) ^n (1/k)=ln n+γ+o_(n∞) (1)

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

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