Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 52349 by Tawa1 last updated on 06/Jan/19

Determine if the series converges or diverges. (i) Σ_(n = 2) ^∞ (1/(n^2 − 1)) (ii) Σ_(n = 1) ^∞ (1/3^(n − 1) )

$$\mathrm{Determine}\:\mathrm{if}\:\mathrm{the}\:\mathrm{series}\:\mathrm{converges}\:\mathrm{or}\:\mathrm{diverges}. \\ $$$$\:\:\:\left(\mathrm{i}\right)\:\:\:\:\underset{\mathrm{n}\:=\:\mathrm{2}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} \:−\:\mathrm{1}} \\ $$$$ \\ $$$$\:\:\:\left(\mathrm{ii}\right)\:\:\:\:\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{n}\:−\:\mathrm{1}} } \\ $$$$\:\:\: \\ $$

Commented by Abdo msup. last updated on 07/Jan/19

1) let S_n =Σ_(k=2) ^n (1/(k^2 −1)) ⇒S_n =(1/2)Σ_(k=2) ^n ((1/(k−1))−(1/(k+1))) =(1/2)Σ_(k=2) ^n (1/(k−1)) −(1/2)Σ_(k=2) ^n (1/(k+1)) but Σ_(k=2) ^n (1/(k−1)) =Σ_(k=1) ^(n−1) (1/k) Σ_(k=2) ^n (1/(k+1)) =Σ_(k=3) ^(n+1) (1/k) =Σ_(k=1) ^(n−1) (1/k) −(3/2) +(1/n) +(1/(n+1)) ⇒S_n =(1/2){Σ_(k=1) ^(n−1) (1/k) −Σ_(k=1) ^(n−1) (1/k) +(3/2)−(1/n)−(1/(n+1))} =(3/4) −(1/(2n)) −(1/(3n+2)) ⇒lim_(n→+∞) S_n =(3/4) .

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

Commented by Abdo msup. last updated on 07/Jan/19

ii)let S_n = Σ_(k=1) ^n (1/3^(k−1) ) =Σ_(k=0) ^(n−1) ((1/3))^k =((1−((1/3))^n )/(1−(1/3))) =(3/2){1−(1/3^n )} ⇒lim_(n→+∞) S_n =(3/2)

$$\left.{ii}\right){let}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{3}^{{k}−\mathrm{1}} }\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{k}} \\ $$$$=\frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}} }{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{{n}} }\right\}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented by Tawa1 last updated on 07/Jan/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jan/19

ii)S_n =(1/3^0 )+(1/3)+(1/3^2 )+...+(1/3^(n−1) ) S_n =((1(1−(1/3^n )))/(1−(1/3)))=((1−(1/3^n ))/(2/3)) S=lim_(n→∞) ((1−(1/3^n ))/(2/3))=((1−0)/(2/3))=(3/2) so converge...

$$\left.{ii}\right){S}_{{n}} =\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{0}} }+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+...+\frac{\mathrm{1}}{\mathrm{3}^{{n}−\mathrm{1}} } \\ $$$${S}_{{n}} =\frac{\mathrm{1}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{{n}} }\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}=\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{{n}} }}{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$${S}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{{n}} }}{\frac{\mathrm{2}}{\mathrm{3}}}=\frac{\mathrm{1}−\mathrm{0}}{\frac{\mathrm{2}}{\mathrm{3}}}=\frac{\mathrm{3}}{\mathrm{2}}\:\:{so}\:{converge}... \\ $$$$ \\ $$

Commented by Tawa1 last updated on 06/Jan/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jan/19

1)T_n =(1/2)[(((n+1)−(n−1))/((n+1)(n−1)))] T_n =(1/2)[(1/(n−1))−(1/(n+1))] T_2 =(1/2)[(1/1)−(1/3)] T_3 =(1/2)[(1/2)−(1/4)] T_4 =(1/2)[(1/3)−(1/5)] T_5 =(1/2)[(1/4)−(1/6)] ... .... now look when we add T_2 +T_3 +T_4 +... all numer cancelled except (1/1) and(1/2) so answdr is=(1/2)[1+(1/2)]=(3/4) so converge...

$$\left.\mathrm{1}\right){T}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\left({n}+\mathrm{1}\right)−\left({n}−\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)}\right] \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right] \\ $$$${T}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}\right] \\ $$$${T}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\right] \\ $$$${T}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}\right] \\ $$$${T}_{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{6}}\right] \\ $$$$... \\ $$$$....\:\:{now}\:{look}\:{when}\:{we}\:{add}\:{T}_{\mathrm{2}} +{T}_{\mathrm{3}} +{T}_{\mathrm{4}} +... \\ $$$${all}\:{numer}\:{cancelled}\:{except}\:\frac{\mathrm{1}}{\mathrm{1}}\:{and}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${so}\:{answdr}\:{is}=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right]=\frac{\mathrm{3}}{\mathrm{4}}\:{so}\:{converge}... \\ $$

Commented by Tawa1 last updated on 06/Jan/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com