Question-194128

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^{\infty}} {(- \frac{1}{2})}^{n} = \underset{n = 0}{\sum^{\infty}} {(- \frac{1}{2})}^{n} - \underset{n = 0}{\sum^{2022}} {(- \frac{1}{2})}^{n}

= \frac{2}{3} - \frac{1 - {(\frac{1}{2})}^{2022}}{\frac{3}{2}} = \frac{1}{3} \times {(\frac{1}{2})}^{2021}

b) \underset{n = 1}{\sum^{\infty}} \frac{22}{n (n + 22)} = \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + 22})

S_{k} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{22} + \dots - \frac{1}{k + 21} - \frac{1}{k + 22}

\underset{n = 1}{\sum^{\infty}} \frac{22}{n (n + 22)} = {l i m}_{k \to \infty} S_{k} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{22}

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