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Question Number 164240 by HongKing last updated on 15/Jan/22

$$\mathrm{Find}:\underset{\mathbf{n}=1}{\sum ^{\mathrm{\infty }}}\ln \left(\mathrm{n}\right)=?$$

Answered by mathmax by abdo last updated on 15/Jan/22

$$\mathrm{this}\mathrm{serie}\mathrm{diverges}\mathrm{to}+\mathrm{\infty }$$

Answered by alephzero last updated on 15/Jan/22

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\ln n=?$$$$\ln 1+\ln 2+...=\ln (1\cdot 2...)=$$$$=\ln \left(\underset{k=1}{\prod ^{\mathrm{\infty }}}k\right)$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\ln n=\ln \left(\underset{k=1}{\prod ^{\mathrm{\infty }}}k\right)$$$$\underset{x\to \mathrm{\infty }}{\lim }\left(\underset{k=1}{\prod ^x}k\right)=\mathrm{\infty }$$$$\underset{x\to \mathrm{\infty }}{\lim }\left(\ln x\right)=\mathrm{\infty }$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\ln n=\mathrm{\infty }$$

Commented by HongKing last updated on 15/Jan/22

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{dear}\mathrm{Sir}$$

Commented by alephzero last updated on 16/Jan/22

$$\mathrm{You}\mathrm{are}\mathrm{welcome},\mathrm{Sir}.$$

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