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S-1-n-1-2n-2-1-4n-4-1-8n-8-S-1-n-i-1-1-2-i-n-2-i-Solvable-




Question Number 7999 by FilupSmith last updated on 27/Sep/16
S=(1/n)+(1/(2n^2 ))+(1/(4n^4 ))+(1/(8n^8 ))+...  S=(1/n)+Σ_(i=1) ^∞ (1/(2^i n^2^i  ))  Solvable?
$${S}=\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{8}{n}^{\mathrm{8}} }+… \\ $$$${S}=\frac{\mathrm{1}}{{n}}+\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{i}} {n}^{\mathrm{2}^{{i}} } } \\ $$$$\mathrm{Solvable}? \\ $$
Answered by Yozzia last updated on 27/Sep/16
Σ_(i=1) ^∞ (1/(2^i n^2^i  ))=Σ_(i=1) ^∞ (1/2^i )((1/n))^2^i      (n∈N)  Σ_(i=1) ^n 2^(−i) x^2^i  =u(x)⇒u′(x)=Σ_(i=1) ^n 2^(−i) ×2^i x^(2^i −1)   u′(x)=x^(−1) Σ_(i=1) ^n x^2^i  =x^(−1) (x^2 +x^4 +x^8 +x^(16) +...+x^2^(n−1)  +x^2^n  )  u′(x)=x^(−1) x^2 (1+x^2 +x^6 +x^(12) +...+x^(2^(n−1) −2) +x^(2^n −2) )  u′(x)=x(1+x^2 +x^6 +x^(12) +...+x^(2^(n−1) −2) +x^(2^n −2) )  {Continue}
$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{i}} {n}^{\mathrm{2}^{{i}} } }=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{i}} }\left(\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2}^{{i}} } \:\:\:\:\left({n}\in\mathbb{N}\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}^{−{i}} {x}^{\mathrm{2}^{{i}} } ={u}\left({x}\right)\Rightarrow{u}'\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}^{−{i}} ×\mathrm{2}^{{i}} {x}^{\mathrm{2}^{{i}} −\mathrm{1}} \\ $$$${u}'\left({x}\right)={x}^{−\mathrm{1}} \underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{x}^{\mathrm{2}^{{i}} } ={x}^{−\mathrm{1}} \left({x}^{\mathrm{2}} +{x}^{\mathrm{4}} +{x}^{\mathrm{8}} +{x}^{\mathrm{16}} +…+{x}^{\mathrm{2}^{{n}−\mathrm{1}} } +{x}^{\mathrm{2}^{{n}} } \right) \\ $$$${u}'\left({x}\right)={x}^{−\mathrm{1}} {x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{6}} +{x}^{\mathrm{12}} +…+{x}^{\mathrm{2}^{{n}−\mathrm{1}} −\mathrm{2}} +{x}^{\mathrm{2}^{{n}} −\mathrm{2}} \right) \\ $$$${u}'\left({x}\right)={x}\left(\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{6}} +{x}^{\mathrm{12}} +…+{x}^{\mathrm{2}^{{n}−\mathrm{1}} −\mathrm{2}} +{x}^{\mathrm{2}^{{n}} −\mathrm{2}} \right) \\ $$$$\left\{{Continue}\right\} \\ $$$$ \\ $$

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