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Question Number 209318 by essaad last updated on 06/Jul/24

Answered by Frix last updated on 06/Jul/24

$$\mathrm{We}\mathrm{know}\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^n}\frac{1}{k}=\mathrm{\infty }$$$$k>1:\frac{1}{\sqrt{k}}>\frac{1}{k}\Rightarrow \underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\sqrt{k}}>\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^n}\frac{1}{\sqrt{k}}=\mathrm{\infty }$$

Answered by mathzup last updated on 07/Jul/24

$$\sum _{k=1}^n\frac{1}{\sqrt{k}}\sim 2\sqrt{n}(n\to +\mathrm{\infty })donclaserie$$$$tendvers+\mathrm{\infty }caddivergente.$$

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