Find-the-sum-of-n-terms-of-the-following-series-1-1-x-1-1-x-1-1-x-

Question Number 6329 by Rasheed Soomro last updated on 24/Jun/16

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{x}}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{x}}}+…. \\ $$

Commented by FilupSmith last updated on 24/Jun/16

sequence x, (√x), x... can be given by: (x/2)(1−(−1)^(n+1) )+((√x)/2)(1−(−1)^n ) 1, 0, 1, 0, ... is given by: (1/2)(1−(−1)^n ) S=Σ_(n=1) ^n ((1/(1+(1/2)(x(1−(−1)^(n+1) )+(√x)(1−(−1)^n ))))) Possibly a simpler method by using an exponential notation. i.e. S=Σ(1/(1+x^a_n )) where a_n toggles between 1 and 1/2 EDIT: Another form: S=Σ_(n=1) ^n (1/(1+x^((1/2)((1−(−1)^(n+1) )+(1/2)(1−(−1)^n ))) ))

$${s}\mathrm{equence}\:{x},\:\sqrt{{x}},\:{x}…\:\mathrm{can}\:\mathrm{be}\:\mathrm{given}\:\mathrm{by}: \\ $$$$\frac{{x}}{\mathrm{2}}\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \right)+\frac{\sqrt{{x}}}{\mathrm{2}}\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \right) \\ $$$$\mathrm{1},\:\mathrm{0},\:\mathrm{1},\:\mathrm{0},\:…\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \right) \\ $$$$ \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left({x}\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \right)+\sqrt{{x}}\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \right)\right)}\right) \\ $$$$\mathrm{Possibly}\:\mathrm{a}\:\mathrm{simpler}\:\mathrm{method}\:\mathrm{by}\:\mathrm{using}\:\mathrm{an} \\ $$$$\mathrm{exponential}\:\mathrm{notation}. \\ $$$$\mathrm{i}.\mathrm{e}.\:{S}=\Sigma\frac{\mathrm{1}}{\mathrm{1}+{x}^{{a}_{{n}} } } \\ $$$$\mathrm{where}\:{a}_{{n}} \:\mathrm{toggles}\:\mathrm{between}\:\mathrm{1}\:\mathrm{and}\:\mathrm{1}/\mathrm{2} \\ $$$$ \\ $$$$\mathrm{EDIT}: \\ $$$$\mathrm{Another}\:\mathrm{form}: \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{1}+{x}^{\frac{\mathrm{1}}{\mathrm{2}}\left(\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \right)\right)} } \\ $$

Commented by FilupSmith last updated on 24/Jun/16

another form: S=Σ_(n=1) ^n (1/(1+x^((1/2)(1−(−1)^(n+1) )) +x^((1/4)(1−(−1)^n )) ))

$$\mathrm{another}\:\mathrm{form}: \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{1}+{x}^{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \right)} +{x}^{\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \right)} } \\ $$

Commented by Rasheed Soomro last updated on 24/Jun/16

$${The}\:{question}\:{is}\:{from}\:{a}\:{text}\:{book} \\ $$$${of}\:{mathematics}\:{and}\:{there}\:{are} \\ $$$${only}\:{three}\:{terms}\:\:{given}. \\ $$

Commented by Yozzii last updated on 24/Jun/16

Are those the only first terms given by the question? pattern on (√x)→−x→−(√x)...?

$${Are}\:{those}\:{the}\:{only}\:{first}\:{terms}\:{given}\:{by} \\ $$$${the}\:{question}? \\ $$$${pattern}\:{on}\:\sqrt{{x}}\rightarrow−{x}\rightarrow−\sqrt{{x}}…? \\ $$

Commented by prakash jain last updated on 24/Jun/16

(1/(1−x))−(1/(1+(√x)))=((1−(1−(√x)))/(1−x))=((√x)/(1−x)) (1/(1−(√x)))−(1/(1−x))=((1+(√x)−1)/(1−x))=((√x)/(1−x)) The given series an AP with common difference ((√x)/(1−x))

$$\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{1}+\sqrt{{x}}}=\frac{\mathrm{1}−\left(\mathrm{1}−\sqrt{{x}}\right)}{\mathrm{1}−{x}}=\frac{\sqrt{{x}}}{\mathrm{1}−{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−\sqrt{{x}}}−\frac{\mathrm{1}}{\mathrm{1}−{x}}=\frac{\mathrm{1}+\sqrt{{x}}−\mathrm{1}}{\mathrm{1}−{x}}=\frac{\sqrt{{x}}}{\mathrm{1}−{x}} \\ $$$$\mathrm{The}\:\mathrm{given}\:\mathrm{series}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{with}\:\mathrm{common}\:\mathrm{difference} \\ $$$$\frac{\sqrt{{x}}}{\mathrm{1}−{x}} \\ $$

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