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Question Number 94556 by ar247 last updated on 19/May/20

Commented by ar247 last updated on 19/May/20

$$help$$

Answered by mr W last updated on 19/May/20

$$\underset{k=0}{\sum ^{\mathrm{\infty }}}{x}^k=\frac{1}{1-x}$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{2018}^k}=\frac{1}{1-\frac{1}{2018}}=\frac{2018}{2017}$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}{x}^k=\frac{1}{1-x}$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}{kx}^{k-1}=\frac{1}{{(1-x)}^2}$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}{kx}^k=\frac{x}{{(1-x)}^2}$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{k}{{2018}^k}=\frac{1}{2018}\times \frac{1}{{(1-\frac{1}{2018})}^2}=\frac{2018}{{2017}^2}$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{2k+1}{{2018}^k}=2\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{k}{{2018}^k}+\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{2018}^k}$$$$=2\times \frac{2018}{{2017}^2}+\frac{2018}{2017}$$$$=\frac{2018\times 2019}{{2017}^2}\approx 1.001487$$

Answered by abdomathmax last updated on 19/May/20

$$\mathrm{S}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{2\mathrm{n}+1}{{2018}^{\mathrm{n}}}\mathrm{let}\mathrm{s}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(2\mathrm{n}+1){\mathrm{x}}^{\mathrm{n}}(\mid \mathrm{x}\mid <1)$$$$\Rightarrow \mathrm{s}\left(\mathrm{x}\right)=2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}=\frac{1}{1-\mathrm{x}}\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}-1}=\frac{1}{{(1-\mathrm{x})}^2}\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}=\frac{\mathrm{x}}{{(1-\mathrm{x})}^2}\Rightarrow \mathrm{s}\left(\mathrm{x}\right)=\frac{2\mathrm{x}}{{(1-\mathrm{x})}^2}+\frac{1}{1-\mathrm{x}}$$$$\mathrm{S}=\mathrm{s}\left(\frac{1}{2018}\right)=\frac{2}{2018{(1-\frac{1}{2018})}^2}+\frac{1}{1-\frac{1}{2018}}$$$$=\frac{2}{2018\times \frac{{2017}^2}{{2018}^2}}+\frac{2018}{2017}$$$$=\frac{2\times 2018}{{2017}^2}+\frac{2018}{2017}=\frac{2\times 2018+2017\times 2018}{{2017}^2}$$$$$$

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