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Question Number 144877 by imjagoll last updated on 30/Jun/21

$$\mathrm{u}+\sqrt{\mathrm{u}}+\sqrt[3]{\mathrm{u}}+\sqrt[4]{\mathrm{u}}+\sqrt[5]{\mathrm{u}}+...+\mathrm{\infty }=?$$$$$$

Answered by MJS_new last updated on 30/Jun/21

$$\mathrm{\forall }u>0:\underset{n=1}{\sum ^{\mathrm{\infty }}}{u}^{1/n}=+\mathrm{\infty }$$$$\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{because}\underset{n\to +\mathrm{\infty }}{\lim }{u}^{1/n}=1$$

Commented by imjagoll last updated on 30/Jun/21

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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