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Question Number 30123 by tawa tawa last updated on 16/Feb/18

$$\mathrm{find}\mathrm{the}\mathrm{convergence}\mathrm{or}\mathrm{divergence}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right)$$

Commented by prof Abdo imad last updated on 16/Feb/18

$$letput{S}_n=\sum _{k=1}^n\frac{k-1}{k}wehave$$$$S_n=n-\sum _{k=1}^n\frac{1}{k}=n-H_{n}but{H}_n=ln\left(n\right)+\gamma +o\left(1\right)$$$$\Rightarrow S_n=n-ln\left(n\right)-\gamma +o\left(1\right)but$$$${lim}_{n\to \mathrm{\infty }}n-ln\left(n\right)={lim}_{n\to \mathrm{\infty }}n(1-\frac{ln\left(n\right)}{n})=+\mathrm{\infty }$$$${lim}_{n\to \mathrm{\infty }}{S}_n=+\mathrm{\infty }andfromthattheserieis$$$$divergente.$$

Commented by tawa tawa last updated on 17/Feb/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by abdo imad last updated on 17/Feb/18

$$manythankssirtawatawa...$$

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