n-0-n-2-n-

Question Number 3734 by prakash jain last updated on 19/Dec/15

$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{n}}{\mathrm{2}^{{n}} }=? \\ $$

Answered by prakash jain last updated on 19/Dec/15

a_n =(n/2^n ), a_(n−1) =((n−1)/2^(n−1) ) lim_(n→∞) (a_n /a_(n−1) )=(n/(n−1))×(1/2)=(1/2) (converges) S=(1/2)+(2/2^2 )+(3/2^3 )+(4/2^4 )+... (A) (S/2)= (1/2^2 )+(2/2^3 )+(3/2^4 )+... (B) A−B (S/2)=(1/2)+(1/2^2 )+(1/2^3 )+(1/2^4 )+... (S/2)=1⇒S=2

$${a}_{{n}} =\frac{{n}}{\mathrm{2}^{{n}} },\:{a}_{{n}−\mathrm{1}} =\frac{{n}−\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{a}_{{n}} }{{a}_{{n}−\mathrm{1}} }=\frac{{n}}{{n}−\mathrm{1}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\:\left({converges}\right) \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{2}^{\mathrm{4}} }+…\:\:\:\left(\mathrm{A}\right) \\ $$$$\frac{{S}}{\mathrm{2}}=\:\:\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{4}} }+…\:\:\:\:\:\:\:\:\left(\mathrm{B}\right) \\ $$$$\mathrm{A}−\mathrm{B} \\ $$$$\frac{{S}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }+… \\ $$$$\frac{{S}}{\mathrm{2}}=\mathrm{1}\Rightarrow{S}=\mathrm{2} \\ $$

Commented by Rasheed Soomro last updated on 19/Dec/15

$$\boldsymbol{\mathcal{T}{han}\mathcal{K}}^{\mathcal{S}} ! \\ $$

Commented by Rasheed Soomro last updated on 19/Dec/15

{ (((S/2)= (1/2^2 )+(2/2^3 )+(3/2^4 )+...)),(((S/2)=(1/2)+(1/2^2 )+(1/2^3 )+(1/2^4 )+...)) :}How is derived?

$$\begin{cases}{\frac{{S}}{\mathrm{2}}=\:\:\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{4}} }+…}\\{\frac{{S}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }+…}\end{cases}\mathcal{H}{ow}\:{is}\:{derived}? \\ $$

Commented by prakash jain last updated on 19/Dec/15

Added clarification. Actually I was working on Q3716 and was trying some series for comparision so posted those.

$$\mathrm{Added}\:\mathrm{clarification}. \\ $$$$\mathrm{Actually}\:\mathrm{I}\:\mathrm{was}\:\mathrm{working}\:\mathrm{on}\:\mathrm{Q3716}\:\mathrm{and}\:\mathrm{was} \\ $$$$\mathrm{trying}\:\mathrm{some}\:\mathrm{series}\:\mathrm{for}\:\mathrm{comparision}\:\mathrm{so} \\ $$$$\mathrm{posted}\:\mathrm{those}. \\ $$

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