Question Number 219025 by universe last updated on 18/Apr/25
Answered by vnm last updated on 19/Apr/25
$${the}\:{generating}\:{function}\:{of}\: \\ $$$${this}\:{sequence}\:{is} \\ $$$${f}\left({x}\right)=\frac{{x}−\mathrm{2ln}\left(\mathrm{1}−{x}\right)+\frac{\mathrm{1}}{\mathrm{1}−{x}}−\mathrm{1}}{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} } \\ $$
Commented by universe last updated on 24/Apr/25
$${sir}\:{can}\:{u}\:{explain}\:{how}\:{to}\:{find}\:\:{f}\left({x}\right) \\ $$
Answered by SdC355 last updated on 19/Apr/25
$${a}_{{n}} =−\left({n}+\mathrm{1}\right)\boldsymbol{\Phi}\left(\mathrm{2},\mathrm{1},{n}+\mathrm{1}\right)+\frac{\boldsymbol{{i}}\pi\left({n}+\mathrm{1}\right)}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$
Answered by vnm last updated on 21/Apr/25
$${f}\left({x}\right)\:{is}\:{a}\:{solution}\:{to}\:{the}\:{differential}\:{equation} \\ $$$$\frac{\mathrm{d}}{\mathrm{d}{x}}{f}\left({x}\right)=\mathrm{2}{f}\left({x}\right)\frac{{x}−\mathrm{1}}{{x}\left(\mathrm{2}−{x}\right)}+\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${f}\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}^{{k}} }{{k}}\right){x}^{{n}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\frac{\mathrm{1}−{t}^{{n}} }{\mathrm{1}−{t}}\mathrm{d}{t}\right){x}^{{n}} \\ $$$${a}_{{n}} =\frac{{n}+\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}^{{k}} }{{k}}=\frac{{n}+\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\frac{\mathrm{1}−{t}^{{n}} }{\mathrm{1}−{t}}\mathrm{d}{t} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} =\mathrm{1} \\ $$