Menu Close

Question-24549

Tinku Tara 0 Comments
Pin




Question Number 24549 by math solver last updated on 20/Nov/17
Commented by math solver last updated on 20/Nov/17
the value of ratio is ??
$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{ratio}\:\mathrm{is}\:?? \\ $$
Commented by prakash jain last updated on 20/Nov/17
1+(1/2^2 )+(1/3^2 )+(1/4^2 )+...=S (1+(1/3^2 )+(1/5^2 )+..)+(1/2^2 )(1+(1/2^2 )+(1/3^2 )+...)=S (1+(1/3^2 )+(1/5^2 )+..)+(S/4)=S (1+(1/3^2 )+(1/5^2 )+..)=((3S)/4) Given ratio (S/(((3S)/4)−(S/4)))=2
$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+…=\mathrm{S} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+..\right)+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…\right)=\mathrm{S} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+..\right)+\frac{\mathrm{S}}{\mathrm{4}}=\mathrm{S} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+..\right)=\frac{\mathrm{3S}}{\mathrm{4}} \\ $$$$\mathrm{Given}\:\mathrm{ratio} \\ $$$$\frac{\mathrm{S}}{\frac{\mathrm{3S}}{\mathrm{4}}−\frac{\mathrm{S}}{\mathrm{4}}}=\mathrm{2} \\ $$
Commented by prakash jain last updated on 20/Nov/17
Terms can be rearranged for absolutely convergent series.
$$\mathrm{Terms}\:\mathrm{can}\:\mathrm{be}\:\mathrm{rearranged}\:\mathrm{for} \\ $$$$\mathrm{absolutely}\:\mathrm{convergent}\:\mathrm{series}. \\ $$
Commented by math solver last updated on 20/Nov/17
god bless you sir .
$$\mathrm{god}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}\:. \\ $$
Commented by maxmathsup by imad last updated on 24/May/19
let q =((1+(1/2^2 ) +(1/3^2 ) +...)/(1−(1/2^2 ) +(1/3^2 ) +..)) ⇒q =((Σ_(n=1) ^∞ (1/n^2 ))/(Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ))) we have Σ_(n=1) ^∞ (1/n^2 ) =(π^2 /6) Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=1) ^∞ (1/((2n)^2 )) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) =−(1/4) (π^2 /6) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) Σ_(n=1) ^∞ (1/n^2 ) =(1/4) Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(3/4).(π^2 /6) =(π^2 /8) ⇒ Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =(π^2 /8) −(π^2 /(24)) =(π^2 /(12)) ⇒ q =((π^2 /6)/(π^2 /(12))) ⇒q =2 .
$${let}\:{q}\:=\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+…}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+..}\:\Rightarrow{q}\:=\frac{\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }}{\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }} \\ $$$${we}\:{have}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\mathrm{4}}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\Rightarrow\:{q}\:=\frac{\frac{\pi^{\mathrm{2}} }{\mathrm{6}}}{\frac{\pi^{\mathrm{2}} }{\mathrm{12}}}\:\Rightarrow{q}\:=\mathrm{2}\:. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *