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Question Number 141149 by Dan last updated on 16/May/21

$${lim}_{x\to \mathrm{\infty }}\underset{i=0}{\sum ^{x-1}}\frac{x}{(x+i)}$$

Answered by mr W last updated on 16/May/21

$$\frac{x}{x+x}<\frac{x}{x+i}<\frac{x}{x}$$$$\frac{1}{2}<\frac{x}{x+i}<1$$$$\frac{x-1}{2}<\underset{i=0}{\sum ^{x-1}}\frac{x}{x+i}<x-1$$$$\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{x-1}{2}\right)<\underset{x\to \mathrm{\infty }}{\lim }\underset{i=0}{\sum ^{x-1}}\frac{x}{x+i}<\underset{x\to \mathrm{\infty }}{\lim }(x-1)$$$$\mathrm{\infty }<\underset{x\to \mathrm{\infty }}{\lim }\underset{i=0}{\sum ^{x-1}}\frac{x}{x+i}<\mathrm{\infty }$$$$\underset{x\to \mathrm{\infty }}{\lim }\underset{i=0}{\sum ^{x-1}}\frac{x}{x+i}=\mathrm{\infty }$$

Answered by Ankushkumarparcha last updated on 16/May/21

$$Solution:Using{L}^{\prime }{Hospital}^{\prime }sRule.$$$$\underset{x\to \mathrm{\infty }}{\lim }\underset{i=0}{\sum ^{i=x-1}}\frac{\frac{d}{dx}x}{\frac{d}{dx}(x+i)}=\underset{x\to \mathrm{\infty }}{\lim }\underset{i=0}{\sum ^{i=x-1}}1(\therefore byputtinglimits)$$$$\underset{x\to \mathrm{\infty }}{\lim }\underset{i=0}{\sum ^{i=x-1}}\frac{x}{x+i}=\mathrm{\infty }$$

Answered by mathmax by abdo last updated on 16/May/21

$$\sum _{\mathrm{i}=0}^{\mathrm{n}-1}\frac{\mathrm{n}}{\mathrm{n}+\mathrm{i}}=\sum _{\mathrm{i}=0}^{\mathrm{n}-1}\frac{1}{1+\frac{\mathrm{i}}{\mathrm{n}}}=\mathrm{n}\times \frac{1}{\mathrm{n}}\sum _{\mathrm{i}=0}^{\mathrm{n}-1}\frac{1}{1+\frac{\mathrm{i}}{\mathrm{n}}}$$$$\mathrm{but}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{1}{\mathrm{n}}\sum _{\mathrm{i}=0}^{\mathrm{n}-1}\frac{1}{1+\frac{\mathrm{i}}{\mathrm{n}}}={\int }_0^1\frac{\mathrm{dx}}{1+\mathrm{x}}=\ln 2\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\sum _{\mathrm{i}=0}^{\mathrm{n}-1}\frac{\mathrm{n}}{\mathrm{n}+\mathrm{i}}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}\left(\mathrm{nlog}2\right)=+\mathrm{\infty }$$$$$$

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