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Question Number 141149 by Dan last updated on 16/May/21
lim_(x→∞) Σ_(i=0) ^(x−1) (x/((x+i)))
$${lim}_{{x}\rightarrow\infty} \underset{{i}=\mathrm{0}} {\overset{{x}−\mathrm{1}} {\sum}}\frac{{x}}{\left({x}+{i}\right)} \\ $$
Answered by mr W last updated on 16/May/21
(x/(x+x))<(x/(x+i))<(x/x) (1/2)<(x/(x+i))<1 ((x−1)/2)<Σ_(i=0) ^(x−1) (x/(x+i))<x−1 lim_(x→∞) (((x−1)/2))<lim_(x→∞) Σ_(i=0) ^(x−1) (x/(x+i))<lim_(x→∞) (x−1) ∞<lim_(x→∞) Σ_(i=0) ^(x−1) (x/(x+i))<∞ lim_(x→∞) Σ_(i=0) ^(x−1) (x/(x+i))=∞
$$\frac{{x}}{{x}+{x}}<\frac{{x}}{{x}+{i}}<\frac{{x}}{{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}<\frac{{x}}{{x}+{i}}<\mathrm{1} \\ $$$$\frac{{x}−\mathrm{1}}{\mathrm{2}}<\underset{{i}=\mathrm{0}} {\overset{{x}−\mathrm{1}} {\sum}}\frac{{x}}{{x}+{i}}<{x}−\mathrm{1} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}−\mathrm{1}}{\mathrm{2}}\right)<\underset{{x}\rightarrow\infty} {\mathrm{lim}}\underset{{i}=\mathrm{0}} {\overset{{x}−\mathrm{1}} {\sum}}\frac{{x}}{{x}+{i}}<\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({x}−\mathrm{1}\right) \\ $$$$\infty<\underset{{x}\rightarrow\infty} {\mathrm{lim}}\underset{{i}=\mathrm{0}} {\overset{{x}−\mathrm{1}} {\sum}}\frac{{x}}{{x}+{i}}<\infty \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\underset{{i}=\mathrm{0}} {\overset{{x}−\mathrm{1}} {\sum}}\frac{{x}}{{x}+{i}}=\infty \\ $$
Answered by Ankushkumarparcha last updated on 16/May/21
Solution: Using L′Hospital′s Rule. lim_(x→∞) Σ_(i=0) ^(i=x−1) (((d/dx) x)/((d/dx) (x+i))) = lim_(x→∞) Σ_(i=0) ^(i=x−1) 1 (∴ by putting limits) lim_(x→∞) Σ_(i=0) ^(i=x−1) (x/(x+i)) = ∞
$${Solution}:\:{Using}\:{L}'{Hospital}'{s}\:{Rule}. \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\underset{{i}=\mathrm{0}} {\overset{{i}={x}−\mathrm{1}} {\sum}}\frac{\frac{{d}}{{dx}}\:{x}}{\frac{{d}}{{dx}}\:\left({x}+{i}\right)}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\underset{{i}=\mathrm{0}} {\overset{{i}={x}−\mathrm{1}} {\sum}}\mathrm{1}\:\:\left(\therefore\:{by}\:{putting}\:{limits}\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\underset{{i}=\mathrm{0}} {\overset{{i}={x}−\mathrm{1}} {\sum}}\frac{{x}}{{x}+{i}}\:=\:\infty \\ $$
Answered by mathmax by abdo last updated on 16/May/21
Σ_(i=0) ^(n−1) (n/(n+i)) =Σ_(i=0) ^(n−1) (1/(1+(i/n))) =n ×(1/n)Σ_(i=0) ^(n−1) (1/(1+(i/n))) but lim_(n→+∞) (1/n)Σ_(i=0) ^(n−1) (1/(1+(i/n)))=∫_0 ^1 (dx/(1+x)) =ln2 ⇒ lim_(n→+∞) Σ_(i=0) ^(n−1) (n/(n+i)) =lim_(n→+∞) (nlog2)=+∞
$$\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{n}}{\mathrm{n}+\mathrm{i}}\:=\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{i}}{\mathrm{n}}}\:=\mathrm{n}\:×\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{i}}{\mathrm{n}}} \\ $$$$\mathrm{but}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{i}}{\mathrm{n}}}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}}\:=\mathrm{ln2}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\:\:\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{n}}{\mathrm{n}+\mathrm{i}}\:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \left(\mathrm{nlog2}\right)=+\infty \\ $$$$ \\ $$

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