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Question Number 70783 by naka3546 last updated on 08/Oct/19

$$\frac{5}{4\cdot 9}+2\left(\frac{9}{16\cdot 25}\right)+3\left(\frac{13}{36\cdot 49}\right)+4\left(\frac{17}{64\cdot 81}\right)+5\left(\frac{21}{100\cdot 121}\right)+\dots =?$$

Commented by tw000001 last updated on 08/Oct/19

$$\mathrm{Well},\mathrm{this}\mathrm{one}\mathrm{is}\mathrm{unconverge}\mathrm{series}\mathrm{because}$$$$\mathrm{difference}\mathrm{and}\mathrm{ratio}\mathrm{are}\mathrm{different}.$$$$\mathrm{But}\mathrm{the}\mathrm{answer}\mathrm{should}\mathrm{be}\mathrm{approximately}$$$$\mathrm{on}\frac{1}{5}\mathrm{with}\mathrm{calculator}.$$

Commented by tw000001 last updated on 08/Oct/19

$$\mathrm{It}\mathrm{can}\mathrm{transform}\mathrm{to}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{4n^2+n}{16n^4+16n^3+4n^2}.$$

Commented by naka3546 last updated on 08/Oct/19

$$anyanswer{that}^{\prime }srealnumberbutnotindecimalnumberform?maybeirrationalnumberform?$$

Commented by prakash jain last updated on 08/Oct/19

$$\frac{4n^2+n}{4n^2{(2n+1)}^2}=\frac{1}{2{(2n+1)}^2}-\frac{1}{2(2n+1)}+\frac{1}{4n}$$$$\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2n+1)}^2}=\frac{1}{2}(\frac{1}{3^2}+\frac{1}{5^2}+..)=\frac{1}{2}(\frac{{\pi }^2}{8}-1)$$$$\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{2n}-\frac{1}{2n+1})=$$$$=\frac{1}{2}(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+..)$$$$=\frac{1}{2}(1-\ln 2)$$$$\mathrm{sum}=\frac{{\pi }^2}{16}-\frac{\ln 2}{2}$$

Answered by mind is power last updated on 08/Oct/19

$$=\sum _{n⩾1}\frac{n(4n+1)}{{\left(2n\right)}^2{(2n+1)}^2}$$$$f\left(x\right)=\frac{x(4x+1)}{4x^2{(2x+1)}^2}=\frac{b}{{(2x+1)}^2}+\frac{c}{x}+\frac{d}{2x+1}$$$$b=\frac{1}{2},$$$$2c+d=0$$$$\frac{1}{18}+c+\frac{d}{3}=\frac{5}{36}$$$$\Rightarrow d+3c=\frac{1}{4}$$$$c=\frac{1}{4},d=\frac{-1}{2}$$$$\frac{x(4x+1)}{4x^2{(2x+1)}^2}=\frac{1}{2{(2x+1)}^2}+\frac{1}{4x}-\frac{1}{2(2x+1)}$$$$\sum _{n⩾1}f\left(n\right)=\mathrm{\Sigma }\frac{1}{2{(2n+1)}^2}+\mathrm{\Sigma }\frac{1}{4n}-\frac{1}{4n+2}$$$$=\frac{1}{2}\mathrm{\Sigma }(\frac{1}{{(2n+1)}^2}+\frac{1}{{\left(2\mathrm{n}\right)}^2}-\frac{1}{{\left(2\mathrm{n}\right)}^2})+\frac{1}{2}\mathrm{\Sigma }\frac{1}{2n}-\frac{1}{2n+1}$$$$=\frac{1}{2}\mathrm{\Sigma }\frac{1}{n^2}-\frac{1}{8}\mathrm{\Sigma }\frac{1}{n^2}+\frac{1}{2}\mathrm{\Sigma }\frac{1}{2n(2n+1)}$$$$=\frac{3}{8}\zeta \left(2\right)+\frac{1}{2}\mathrm{\Sigma }\frac{x^{2n+1}}{2n(2n+1)}\mid x=1$$$$f\left(x\right)=\frac{3}{8}\zeta \left(2\right)+\frac{1}{2}\mathrm{\Sigma }\frac{x^{2n+1}}{2n(2n+1)}$$$$f^{\prime }\left(x\right)=\frac{1}{2}\mathrm{\Sigma }\frac{x^{2n}}{2n}=\frac{1}{4}\sum _{n⩾1}\frac{{\left(x^2\right)}^n}{n}=-\frac{1}{4}ln(1-x^2)$$$$f\left(x\right)=-\frac{1}{4}\int ln(1-x^2)dx$$$$=-\frac{1}{4}xln(1-x^2)+\frac{1}{4}\int \frac{-2x^2}{1-x^2}dx$$$$=-\frac{xln(1-x^2)}{4}-\frac{1}{2}\int -1+\frac{1}{1-x^2}dx$$$$=\frac{-xln(1-x^2)}{4}+\frac{x}{2}-\frac{1}{2}\int \frac{dx}{(1-x)(1+x)}$$$$=\frac{-xln(1-x^2)}{4}+\frac{x}{2}-\frac{1}{4}\int \frac{1}{1+x}dx-\frac{1}{4}\int \frac{1}{1-x}dx$$$$=\frac{-xln(1-x^2)}{4}+\frac{x}{2}+\frac{1}{4}ln\left(\frac{1-x}{1+x}\right)+c$$$$f\left(0\right)=\frac{3}{8}\zeta \left(2\right)$$$$\Rightarrow$$$$f\left(x\right)=\frac{-xln(1-x)-xln(1+x)+2x+ln(1-x)-ln(1+x)}{4}+\frac{3\zeta \left(2\right)}{8}$$$$=\frac{(1-x)ln(1-x)-(1+x)ln(1+x)+2x}{4}+\frac{3\zeta \left(2\right)}{8}$$$$limx\to 1f\left(x\right)=\frac{2-2ln\left(2\right)}{4}+\frac{3\zeta \left(2\right)}{8}=\sum _{n⩾1}f\left(n\right)$$$$\Rightarrow$$$$\sum _{n⩾1}\frac{n(4n+1)}{{\left(2n\right)}^2{(2n+1)}^2}=\frac{2-2ln\left(2\right)}{4}+\frac{3}{8}.\frac{{\pi }^2}{6}$$$$$$

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