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Question Number 194128 by tri26112004 last updated on 28/Jun/23

Answered by MM42 last updated on 28/Jun/23

$$a)\underset{n=2023}{\sum ^{\mathrm{\infty }}}{(-\frac{1}{2})}^n=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-\frac{1}{2})}^n-\underset{n=0}{\sum ^{2022}}{(-\frac{1}{2})}^n$$$$=\frac{2}{3}-\frac{1-{\left(\frac{1}{2}\right)}^{2022}}{\frac{3}{2}}=\frac{1}{3}\times {\left(\frac{1}{2}\right)}^{2021}$$$$b)\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{22}{n(n+22)}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n}-\frac{1}{n+22})$$$$S_k=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{22}+...-\frac{1}{k+21}-\frac{1}{k+22}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{22}{n(n+22)}={lim}_{k\to \mathrm{\infty }}{S}_k=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{22}$$$$$$

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