Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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)=+∞

$$\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{n}}{\mathrm{n}+\mathrm{i}}\:=\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{i}}{\mathrm{n}}}\:=\mathrm{n}\:×\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{i}}{\mathrm{n}}} \\ $$$$\mathrm{but}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{i}}{\mathrm{n}}}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}}\:=\mathrm{ln2}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\:\:\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{n}}{\mathrm{n}+\mathrm{i}}\:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \left(\mathrm{nlog2}\right)=+\infty \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com