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Question Number 14047 by FilupS last updated on 27/May/17

$$S=1+i-1-i+1+...$$$$\frac{1}{i}=-i$$$$S=i(-i+1+i-1-i+1+...)$$$$S=i(-i+S)$$$$S=1+iS$$$$S(1-i)=1$$$$\therefore S=\frac{1}{1-i}$$$$$$$$\mathrm{a})\mathrm{Is}\mathrm{this}\mathrm{correct}?$$$$\mathrm{b})\mathrm{Do}\mathrm{there}\mathrm{exist}\mathrm{any}\mathrm{other}\mathrm{sequences}$$$$\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}:$$$$S=(a_1+...+a_n)+(a_1+...+a_n)+...$$$$S=(a_1+...+a_n)\left(\underset{m\mathrm{times}}{1+1+...+1}\right)$$$$\Rightarrow S=\underset{i=1}{\sum ^{m\to \mathrm{\infty }}}\underset{j=1}{\sum ^n}{a}_j$$$$\mathrm{where}{a}_{t+1}={ba}_t,a_1={ba}_n$$$$$$$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{very}\mathrm{interested}\mathrm{in}\mathrm{these}\mathrm{sequences}$$

Commented by prakash jain last updated on 27/May/17

$$\mathrm{This}\mathrm{is}\mathrm{geometric}\mathrm{series}\mathrm{with}$$$$\mathrm{common}\mathrm{ratio}i.$$$$\mathrm{Since}\mid i\mid =1,\mathrm{by}\mathrm{geometric}\mathrm{series}$$$$\mathrm{test}\mathrm{the}\mathrm{series}\mathrm{diverges}.$$$$S_{4n}=0$$$$S_{4n+1}=1$$$$S_{4n+2}=1+i$$$$S_{4n+3}=i$$$$S=\frac{0+\left(1\right)+(1+i)+\left(i\right)}{4}=\frac{1+i}{2}$$

Commented by prakash jain last updated on 27/May/17

$$S=\frac{1}{1+i}\mathrm{is}\mathrm{not}\mathrm{correct}.\mathrm{You}\mathrm{can}$$$$\mathrm{get}4\mathrm{different}\mathrm{answers}\mathrm{for}S.$$$$S=1+i-1-i+1+i-1-i$$$$=(1+i)-(1+i)+(1+i)+...$$$$=(1+i)-[(1+i)-(1+i)...]$$$$=(1+i)-S$$$$2S=(1+i)\Rightarrow S=\frac{1+i}{2}$$

Commented by RasheedSindhi last updated on 27/May/17

$$S=1+iS$$$$\stackrel{?}{\Rightarrow }S(1+i)=1$$$$$$

Commented by ajfour last updated on 27/May/17

$$S_n=(\frac{1}{2}+\frac{i}{2})(1-i^n)$$$$S_{4n}=0$$$$S_{4n+1}=1$$$$S_{4n+2}=1+i$$$$S_{4n+3}=i.$$

Commented by ajfour last updated on 27/May/17

Commented by prakash jain last updated on 27/May/17

$$S_n\mathrm{is}\mathrm{fine}\mathrm{geometric}\mathrm{series}$$$$\mathrm{and}\mathrm{finite}\mathrm{number}\mathrm{of}\mathrm{terms}$$$$\mathrm{converges}.$$$$\mathrm{Geometric}\mathrm{series}\mathrm{with}\mid r\mid ⩾1$$$$\mathrm{diverges}.$$$$\mathrm{For}\mathrm{example}$$$$\mathrm{S}=1-1+1-1+...$$$$\mathrm{diverges}\mathrm{and}\mathrm{does}\mathrm{not}\mathrm{have}\mathrm{a}\mathrm{fixed}$$$$\mathrm{sum}.$$$$\mathrm{S}=1or0$$$$\mathrm{so}S\mathrm{can}\mathrm{be}\mathrm{taken}\mathrm{as}\frac{1}{2}\mathrm{using}$$$$\mathrm{analytical}\mathrm{continuation}.$$

Commented by FilupS last updated on 27/May/17

$$S=\frac{1}{1+i}\mathrm{has}\mathrm{been}\mathrm{corrected}\mathrm{to}S=\frac{1}{1-i}$$$$$$$$\mathrm{This}\mathrm{is}\mathrm{all}\mathrm{interesting}.$$$$\mathrm{I}\mathrm{am}\mathrm{aware}\mathrm{of}S=1-1+1-1+...$$$$\mathrm{These}\mathrm{kinds}\mathrm{if}\mathrm{sequences}\mathrm{are}\mathrm{interesting}$$

Commented by prakash jain last updated on 27/May/17

$$\mathrm{For}\mathrm{the}\mathrm{sequence}\mathrm{of}\mathrm{the}\mathrm{type}.$$$$a_{t+1}={na}_t,a_1={ba}_n$$$$a_n=a_1{b}^{n-1}$$$$a_1={ba}_n$$$$\Rightarrow b^n=1$$$$b^n=1=e^{2\pi i}$$$$b=e^{2\pi ki/n},k\in \{0,1,2,...,n-1\}$$$$b\mathrm{is}{n}^{th}\mathrm{root}\mathrm{of}\mathrm{unity}.$$

Commented by ajfour last updated on 27/May/17

$$if\mathrm{S}=1-1+1-1+...$$$$thenevenwehaveaformula$$$$for{\mathrm{S}}_n:$$$$S_n=\frac{1}{2}(1-i^{2n}).$$

Commented by prakash jain last updated on 27/May/17

$$\frac{1}{1-i}=\frac{1+i}{2}\mathrm{is}\mathrm{correct}.$$$$\mathrm{This}\mathrm{value}\mathrm{is}\mathrm{correct}\mathrm{with}\mathrm{analytic}$$$$\mathrm{continuation}.\mathrm{This}\mathrm{series}\mathrm{itself}$$$$\mathrm{is}\mathrm{divergent}.$$$$\mathrm{S}=1+2+2^2+....$$$$\mathrm{S}=1+2(1+2+2^2+..)$$$$\mathrm{S}=-1.$$$$\mathrm{S}=-1\mathrm{is}\mathrm{derived}\mathrm{from}\mathrm{analytical}$$$$\mathrm{continuation}\mathrm{of}f\left(x\right)=\frac{1}{1-x}$$$$=1+x+x^2+...$$

Commented by ajfour last updated on 27/May/17

$$incomprehensibleindeed!$$

Commented by prakash jain last updated on 27/May/17

$$\mathrm{Analytical}\mathrm{continuation}\mathrm{is}\mathrm{interesting}$$$$\mathrm{for}\mathrm{example}\mathrm{using}\mathrm{analytical}$$$$\mathrm{continuation}\mathrm{it}\mathrm{can}\mathrm{be}\mathrm{proved}$$$$\mathrm{that}$$$$1+2+3+4+...\mathrm{to}\mathrm{infinity}=-\frac{1}{12}$$

Commented by FilupS last updated on 27/May/17

$$\mathrm{Yeah},\mathrm{i}\mathrm{know}\mathrm{a}\mathrm{bit}\mathrm{about}\mathrm{the}\mathrm{topic}.$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{really}\mathrm{see}{}^{\prime }{how}^{\prime }\mathrm{half}\mathrm{of}\mathrm{the}\mathrm{solutions}$$$$\mathrm{are}\mathrm{logical},\mathrm{but}\mathrm{i}\mathrm{love}\mathrm{it}$$

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