Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 144380 by SOMEDAVONG last updated on 25/Jun/21

L=lim_(n→+∝) ((1/(1^2 +2(1))) + (1/(2^2 +2n)) +...+ (1/(n^2 +2n)))=?

$$\mathrm{L}=\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} +\mathrm{2n}}\:+...+\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{2n}}\right)=? \\ $$

Answered by mathmax by abdo last updated on 25/Jun/21

S_n =Σ_(k=1) ^n (1/(k^2 +2k)) ⇒S_n =Σ_(k=1) ^(n ) (1/(k(k+2))) =(1/2)Σ_(k=1) ^n ((1/k)−(1/(k+2)))=(1/2)Σ_(k=1) ^n (1/k)−(1/2)Σ_(k=1) ^n (1/(k+2))(k+2=p) =(1/2)Σ_(k=1) ^n (1/k)−(1/2)Σ_(p=3) ^(n+2) (1/p) =(1/2)(1+(1/2)+Σ_(k=3) ^n (1/k))−(1/2)Σ_(k=3) ^n (1/k)−(1/2)((1/(n+1))+(1/(n+2))) =(3/4)−(1/2)((1/(n+1))+(1/(n+2)))⇒lim_(n→+∞) S_n =(3/4)

$$\mathrm{S}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} \:+\mathrm{2k}}\:\Rightarrow\mathrm{S}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}\:} \:\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{k}+\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left(\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{2}}\left(\mathrm{k}+\mathrm{2}=\mathrm{p}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{p}=\mathrm{3}} ^{\mathrm{n}+\mathrm{2}} \:\frac{\mathrm{1}}{\mathrm{p}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\sum_{\mathrm{k}=\mathrm{3}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{3}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\right)\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{S}_{\mathrm{n}} =\frac{\mathrm{3}}{\mathrm{4}} \\ $$

Commented by SOMEDAVONG last updated on 25/Jun/21

Thanks sir!

$$\mathrm{Thanks}\:\mathrm{sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com