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Is-the-following-correct-S-i-1-2-i-1-S-1-2-4-8-16-32-64-128-2S-2-4-8-6-32-2S-S-1-S-1-

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Question Number 4365 by Filup last updated on 13/Jan/16
Is the following correct? S=Σ_(i=1) ^∞ 2^(i−1) S=1+2+4+8+16+32+64+128+... ∴2S=2+4+8+6+32+... 2S=S−1 S=−1
$$\mathrm{Is}\:\mathrm{the}\:\mathrm{following}\:\mathrm{correct}? \\ $$$$ \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{2}^{{i}−\mathrm{1}} \\ $$$${S}=\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{8}+\mathrm{16}+\mathrm{32}+\mathrm{64}+\mathrm{128}+… \\ $$$$\therefore\mathrm{2}{S}=\mathrm{2}+\mathrm{4}+\mathrm{8}+\mathrm{6}+\mathrm{32}+… \\ $$$$\mathrm{2}{S}={S}−\mathrm{1} \\ $$$${S}=−\mathrm{1} \\ $$
Commented by Filup last updated on 13/Jan/16
or S=Σ_(i=n) ^∞ 2^i S=2^n +2^(n+1) +2^(n+2) +... 2S=2^(n+1) +2^(n+2) +2^(n+3) +... 2S=S−2^n ∴S=−2^n can be taken further: S=Σ_(i=n) ^∞ a^i S=a^n +a^(n+1) +a^(n+2) +... aS=a^(n+1) +a^(n+2) +a^(n+3) +... aS=S−a^n S=−(a^n /(a−1)) ∴S=(a^n /(1−a))
$${or} \\ $$$${S}=\underset{{i}={n}} {\overset{\infty} {\sum}}\mathrm{2}^{{i}} \\ $$$${S}=\mathrm{2}^{{n}} +\mathrm{2}^{{n}+\mathrm{1}} +\mathrm{2}^{{n}+\mathrm{2}} +… \\ $$$$\mathrm{2}{S}=\mathrm{2}^{{n}+\mathrm{1}} +\mathrm{2}^{{n}+\mathrm{2}} +\mathrm{2}^{{n}+\mathrm{3}} +… \\ $$$$\mathrm{2}{S}={S}−\mathrm{2}^{{n}} \\ $$$$\therefore{S}=−\mathrm{2}^{{n}} \\ $$$$ \\ $$$${can}\:{be}\:{taken}\:{further}: \\ $$$${S}=\underset{{i}={n}} {\overset{\infty} {\sum}}{a}^{{i}} \\ $$$${S}={a}^{{n}} +{a}^{{n}+\mathrm{1}} +{a}^{{n}+\mathrm{2}} +… \\ $$$${aS}={a}^{{n}+\mathrm{1}} +{a}^{{n}+\mathrm{2}} +{a}^{{n}+\mathrm{3}} +… \\ $$$${aS}={S}−{a}^{{n}} \\ $$$${S}=−\frac{{a}^{{n}} }{{a}−\mathrm{1}} \\ $$$$\therefore{S}=\frac{{a}^{{n}} }{\mathrm{1}−{a}} \\ $$
Commented by Filup last updated on 13/Jan/16
is S=(a^n /(1−a)) only true as a limiting sum? or is it an analytical solution?
$$\mathrm{is}\:\:\:{S}=\frac{{a}^{{n}} }{\mathrm{1}−{a}}\:\:\:\mathrm{only}\:\mathrm{true}\:\mathrm{as}\:\mathrm{a}\:\mathrm{limiting}\:\mathrm{sum}? \\ $$$$\mathrm{or}\:\mathrm{is}\:\mathrm{it}\:\mathrm{an}\:\mathrm{analytical}\:\mathrm{solution}? \\ $$
Commented by Yozzii last updated on 13/Jan/16
S(x)=Σ_(i=1) ^∞ x^(i−1) =1+x+x^2 +x^3 +... xS(x)=x+x^2 +x^3 +x^4 +... ∴(1−x)S(x)=1 S(x)=(1/(1−x)). This assumes x≠1 and that S(x), being the limit of Σ_(i=1) ^∞ x^(i−1) , exists. Now,let S(x,n)=Σ_(i=1) ^n x^(i−1) =((x^n −1)/(x−1)) (x≠1) S(x,n)=(x^n /(x−1))+(1/(1−x)). If ∣x∣>1⇒lim_(n→∞) x^n does not exist. ⇒lim_(n→∞) S(x,n) does not exist. So, if x=2>1 S does not have a limit so that S≠−1. It′s interesting how you′ve shown S=−1. But by the concept of number as a vector how can the infinite sum of positive numbers (vectors) give a negative number (opposite vector direction)?
$${S}\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{i}−\mathrm{1}} =\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +… \\ $$$${xS}\left({x}\right)={x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +… \\ $$$$\therefore\left(\mathrm{1}−{x}\right){S}\left({x}\right)=\mathrm{1} \\ $$$${S}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}.\:{This}\:{assumes}\:{x}\neq\mathrm{1}\: \\ $$$${and}\:{that}\:{S}\left({x}\right),\:{being}\:{the}\:{limit}\:{of}\:\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{i}−\mathrm{1}} , \\ $$$${exists}.\: \\ $$$${Now},{let}\:{S}\left({x},{n}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{x}^{{i}−\mathrm{1}} =\frac{{x}^{{n}} −\mathrm{1}}{{x}−\mathrm{1}}\:\:\left({x}\neq\mathrm{1}\right) \\ $$$${S}\left({x},{n}\right)=\frac{{x}^{{n}} }{{x}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}−{x}}. \\ $$$${If}\:\mid{x}\mid>\mathrm{1}\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}{x}^{{n}} \:{does}\:{not}\:{exist}.\: \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}{S}\left({x},{n}\right)\:{does}\:{not}\:{exist}.\:{So},\:{if}\:{x}=\mathrm{2}>\mathrm{1} \\ $$$${S}\:{does}\:{not}\:{have}\:{a}\:{limit}\:{so}\:{that}\:{S}\neq−\mathrm{1}. \\ $$$$ \\ $$$${It}'{s}\:{interesting}\:{how}\:{you}'{ve}\:{shown} \\ $$$${S}=−\mathrm{1}.\:{But}\:{by}\:{the}\:{concept}\:{of}\:{number} \\ $$$${as}\:{a}\:{vector}\:{how}\:{can}\:{the}\:{infinite} \\ $$$${sum}\:{of}\:{positive}\:{numbers}\:\left({vectors}\right)\: \\ $$$${give}\:{a}\:{negative}\:{number}\:\left({opposite}\:{vector}\right. \\ $$$$\left.{direction}\right)? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by prakash jain last updated on 13/Jan/16
S=−1 using analytical continuity. f(x)=(1/(1−x))
$${S}=−\mathrm{1}\:\mathrm{using}\:\mathrm{analytical}\:\mathrm{continuity}. \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$
Commented by malwaan last updated on 14/Jan/16
I think s=−1 is wronge
$${I}\:{think}\:{s}=−\mathrm{1}\:{is}\:{wronge} \\ $$

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