Question Number 2941 by Syaka last updated on 30/Nov/15

(1/2) + (3/4) + (5/8) + (7/(16)) + ....... = ?

Commented bySyaka last updated on 01/Dec/15

Thanks for Solution Sir Rasheed and also for Solved from Sir Yozzi. Like that.

Answered by Rasheed Soomro last updated on 01/Dec/15

Let S=(1/2) + (3/4) + (5/8) + (7/(16)) + ....... = ? Genral term =((1+(n−1)×2)/2^n )=((2n−1)/2^n )=((2n)/2^n )−(1/2^n )=(n/2^(n−1) )−(1/2^n ) So, S=Σ_(n=1) ^(∞) ((n/2^(n−1) )−(1/2^n ))=Σ_(n=1) ^(∞) ((n/2^(n−1) )) −Σ_(n=1) ^(∞) ((1/2^n )) S=((1/2^0 )−(1/2^1 ))+((2/2^1 )−(1/2^2 ))+((3/2^2 )−(1/2^3 ))+((4/2^3 )−(1/2^4 ))+... S=((1/2^0 )+(2/2^1 )+(3/2^2 )+(4/2^3 )+...)−((1/2^1 )+(1/2^2 )+(1/2^3 )+(1/2^4 )+...) S=((1/2^0 )+(2/2^1 )+(3/2^2 )+(4/2^3 )+...)−(((1/2)/(1−(1/2)))) S=(1+(2/2^1 )+(3/2^2 )+(4/2^3 )+...)−1 S=(2/2^1 )+(3/2^2 )+(4/2^3 )+...=(1/2)[(2/2^0 )+(3/2^1 )+(4/2^2 )+...] General term: ((n+1)/2^(n−1) )=(n/2^(n−1) )+(1/2^(n−1) ) S=(1/2)[((1/2^1 )+(2/2^2 )+(3/2^3 )+...)+((1/2^0 )+(1/2^1 )+(1/2^2 )+...)] S=(1/2)[((1/2^0 )+(2/2^1 )+(3/2^2 )+...)+(1/(1−(1/2)))] S=(1/2)[((1/2^0 )+(2/2^1 )+(3/2^2 )+...)+2]>(1/2)[((1/2^0 )+(1/2^1 )+(1/2^2 )+...)+2] S>(1/2)[(1/(1−(1/2)))+2]=(1/2)(2+2)=2 S>2 Continue

Commented byYozzi last updated on 01/Dec/15

Let Q=Σ_(n=1) ^N nx^(n−1) =Σ_(n=1) ^N (d/dx)(x^n ) (x∈R) Q=(d/dx)(Σ_(n=1) ^N x^n ) Q=(d/dx)(xΣ_(n=1) ^N x^(n−1) ) Q=(d/dx)(x((x^N −1)/(x−1))) (x≠1) Q=(d/dx)(((x^(N+1) −x)/(x−1))) Q=(((x−1)((N+1)x^N −1)−(1)(x^(N+1) −x))/((x−1)^2 )) Q=(((N+1)x^(N+1) −x−(N+1)x^N +1−x^(N+1) +x)/((x−1)^2 )) Q=((Nx^(N+1) −(N+1)x^N +1)/((x−1)^2 )) Let x=0.5 ∴Q=((N(0.5)^(N+1) −(N+1)(0.5)^N +1)/((0.5−1)^2 )) Q=4(N(0.5)^(N+1) −(N+1)(0.5)^N +1) lim_(N→∞) Q=4{lim_(N→∞) N(0.5)^(N+1) −lim_(N→∞) (N+1)(0.5)^N +1} N(0.5)^(N+1) =(N/2^(N+1) )→(∞/∞) (indeterminate) N→∞. By use of L′Hopital′s rule, lim_(N→∞) (N/2^(N+1) )=lim_(N→∞) (1/(2^(N+1) ln2))=(1/∞)=0 and similarly lim_(N→∞) (N+1)(0.5)^N =0 ∴ lim_(N→∞) Q=4(0−0+1)=4 ⇒S=4−((1/2)/(1−0.5))=3

Commented byRasheedAhmad last updated on 02/Dec/15

Approach that I couldn′t even think of!

Answered by Rasheed Soomro last updated on 03/Dec/15

Let S=(1/2) + (3/4) + (5/8) + (7/(16)) + ...... General Term:((2n−1)/2^n )=(n/2^(n−1) )−(1/2^n ) S=Σ_(n=1) ^(∞) ((n/2^(n−1) )−(1/2^n ))=Σ_(n=1) ^(∞) ( (n/2^(n−1) ))−Σ_(n=1) ^(∞) ((1/2^n )) =Σ_(n=1) ^(∞) (n((1/2))^(n−1) ) −((1/2)/(1−(1/2))) =Σ_(n=1) ^(∞) (n((1/2))^(n−1) ) −1 With the help of Yozzi ′s comment Let (1/2) (written in red)→x For x=(1/2) : S =Σ_(n=1) ^(∞) (nx^(n−1) )−1 S=−1+Σ_(n=1) ^(∞) (d/dx)(x^n ) S=−1+(d/dx)(lim_(N→∞) Σ_(n=1) ^(N) (x^n )) =−1+lim_(N→∞) [(d/dx)(((x(x^N −1))/(x−1)))] =−1+lim_(N→∞) [(d/dx)(((x^(N+1) −1)/(x−1)))] =−1+lim_(N→∞) [(((x−1)(d/dx)(x^(N+1) −1)−(x^(N+1) −1)(d/dx)(x−1))/((x−1)^2 ))] =−1+lim_(N→∞) [(((x−1)(N+1)x^N −(x^(N+1) −1)(1))/((x−1)^2 ))] =−1+lim_(N→∞) [(((N+1)x^(N+1) −(N+1)x^N −x^(N+1) +1)/((x−1)^2 ))] =−1+lim_(N→∞) [(((N+1)x^(N+1) −x^(N+1) −(N+1)x^N +1)/((x−1)^2 ))] =−1+lim_(N→∞) [(((N+1−1)x^(N+1) −(N+1)x^N +1)/((x−1)^2 ))] =−1+lim_(N→∞) [((Nx^(N+1) −(N+1)x^N +1)/((x−1)^2 ))] ∵ x=(1/2) =−1+lim_(N→∞) [((N((1/2))^(N+1) −(N+1)((1/2))^N +1)/(((1/2)−1)^2 ))] =−1+lim_(N→∞) [((N((1/2))^(N+1) −(N+1)((1/2))^N +1)/(1/4))] =−1+4 lim_(N→∞) [(N/2^(N+1) )−((N+1)/2^N )+1] =−1+4 [lim_(N→∞) (N/2^(N+1) )−lim_(N→∞) ((N+1)/2^N )+lim_(N→∞) 1] By L′Hopital ′s rule lim_(N→∞) (N/2^(N+1) )=0,lim_(N→∞) ((N+1)/2^N )=0 =−1+4(1)=3